# Can you please help me get the solution to this limit without using squeeze theorem and l'hopitals rule lim x to 0 of x^3 sin(1/x) lim x to 0 of x^2 sin^2(1/x)

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**Calculus**

Can you please help me get the solution to this limit without using squeeze theorem and l'hopitals rule lim x to 0 of x^3 sin(1/x) lim x to 0 of x^2 sin^2(1/x)

**calculus**

using the squeeze theorem, find the limit as x->0 of x*e^[8sin(1/x)] what i did was: -1<=sin(1/x)<=1 -8<=8*sin(1/x)<=8 e^(-8)<=e^[8*sin(1/x)]<=e^(8) x*e^(-8)<=x*e^[8*sin(1/x)]<=x*e^(8) lim x->0 [x*e^(-8)] = 0 lim x->0 [x*e^(8)] = 0 hence, lim x...

**maths**

find the limit of lim sin(x-1)/(x^2 + x - 2) x->1 without using l'hopitals rule

**calculus**

Lim sin2h sin3h / h^2 h-->0 how would you do this ?? i got 6 as the answer, just want to make sure it's right. and i couldn't get this one (use theorem 2) lim tanx/x x-->0 and also this one (use squeeze theorem to evaluate the limit) lim (x-1)sin Pi/x-1 x-->1

**Calculus**

Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. 1. Int. of (xlogx)dx from 0 to 1. Indefinite...

**Calculus**

Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. 1. Int. of (xlogx)dx from 0 to 1. Indefinite...

**Calculus**

1.) Find the equation of the line that is tangent to the graph of y-y^3 at x=1. 2.) lim x->0 ((sin x*cos 2x)/3x) 3.) Show, using the squeeze theorem, that the limit of Xe^(sin 1/x) as x->0 is 0.

**L'Hopitals Rule**

5) Use the L’Hopital’s method to evaluate the following limits. In each case, indicate what type of limit it is ( 0/0, ∞/∞, or 0∙∞) lim x→2 sin(x^2−4)/(x−2) = lim x→+∞ ln(x−3)/(x−5) = lim x→pi/4 (x&#...

**calculus**

I am studying for a final exam and in our review packet we have this question, which I am having trouble with: lim (x-->0) ((e^x)+x)^(1/x) I tried this problem two ways. The first was with L'Hopitals rule... lim(x-->0) (1/x)((e^x)+x)^(1/x-1) * (e^x+1) but that doesn't ...

**calculus**

Suppose that f(x) is bounded: that is, there exists a constant M such that abs(f(x)) is < or equal to M for all x. Use the squeeze theorem to prove that lim x^2f(x)=0 as x approaches 0. if g(x) is Mx^2 then what is f(x) and h(x) according to the squeeze theorem. Also, what ...

**calc**

need to find: lim as x -> 0 of 4(e^2x - 1) / (e^x -1) Try splitting the limit for the numerator and denominator lim lim x->0 4(e^2x-1) (4)x->0 (e^2x-1) ______________ = ________________ lim lim x->0 e^X-1 x->0 e^x-1 Next solve for lim x->0 and simplify

**Calculus**

Find the limit. Use L'Hopitals Rule if necessary. lim (x^2+3x+2)/(x^2+x) x -> 0

**Calc Limits**

lim (1+x)^1/x. Give an exact answer. x->0 This reads: The limits as x approaches zero is (1 plus x) to the 1 divided by x. The log of each term is (1/x) ln (1 + x) = ln (1+x)/x Using L'Hopital's rule for the limit of f(x)/g(x), the limit if the log is lim f'/g' = lim [1/(1+...

**Calculus**

L'Hospital's Rule Lim(lnx)^(x-1) x->1+ This is what I did so far... Form: 0^0 ln(lnx)/ (1/x-1) ----------------------------------- I am not sure if you could use L'Hospital Rule (as stated in the instruction)for the 2 problems below: lim [ln(y^2+2y)]/[lny]=1 y->0+ lim ...

**math**

evaluate without using L'Hopital theorem the following limit lim x-->0 [(sin(x)-x)/(x-tan(x))] the answer is 0.5 but I want to know the steps to calculate such a problem

**math**

evaluate without using L'Hopital theorem the following limit lim x-->0 [(sin(x)-x)/(x-tan(x))] the answer is 0.5 but I want to know the steps to calculate such a problem

**calc bc (condensed**

is the limit as x approaches 0 of sin3x over 3x equal to zero? sorry-- basically this is my problem: lim [sin 3x / 4x) x-> 0 ~~~~I multiplied& eventually got to .75* lim (sin 3x / 3x) x-> 0 ~so i figured since (lim (sinx/x) x-> 0 was equal to zero, then lim (sin3x/ 3x...

**Calculus**

yes! tnk u ok? It's actually (x->0.) Find the limit of cot(x)-csc(x) as x approached 0? Lim [cot(x) - csc (x)] ..x->0 = Lim [(cos x -1)/sin x] ..x->0 Use L'Hopital's rule and take the ratio of the derivatives: Lim (-sin x/cos x) = 0 x->0 thank you very much...it ...

**CALCULUS**

Could someone please solve these four problems with explanations? I'd like to understand how to get to the answers. Thank you! Without using a calculator: For each of the following, find: I. lim x->a- f(x) II. lim x->a+ f(x) III. lim x->a f(x) A. f(x)=|x^2+3x+2|/x^2-4...

**calc**

Use L’Hopital’s rule to find the limit of this sequence (n^100)/(e^n) ...If you do L'Hop. Rule it would take forever, right? You would always get an (e^n) at the bottom and will have to use the L'Hop. rule 100 times to find the limit...100*n^99, 9900n^98, and etc. Is there...

**Calculus**

Find the limit. lim 5-x/(x^2-25) x-->5 Here is the work I have so far: lim 5-x/(x^2-25) = lim 5-x/(x-5)(x+5) x-->5 x-->5 lim (1/x+5) = lim 1/10 x-->5 x-->5 I just wanted to double check with someone and see if the answer is supposed to be positive or negative. ...

**Calculus. Limits. Check my answers, please! :)**

4. lim (tanx)= x->pi/3 -(sqrt3) 1 (sqrt3) ***-1 The limit does not exist. 5. lim |x|= x->-2 -2 ***2 0 -1 The limit does not exist. 6. lim [[x]]= x->9/2 (Remember that [[x]] represents the greatest integer function of x.) 4 5 ***4.5 -4 The limit does not exist. 7. lim...

**Calculus**

Use the graph to estimate the limit: lim x->0 sin(3x)/x When x is in degrees lim x->0 sin(3x)/x = ________ I thought the the answer was (3*180)/pi but it's not... please help... Thanks

**calculus**

using L'Hospital's rule, evalutate; LIM as x->0 e^x +cos x / e^x + sin x I'm at LIM x->0 e^x+cosx / e^x +sinx but now I at a lost as to how to proceed.

**CALCULUS - need help!**

Determine the limit of the trigonometric function (if it exists). 1. lim sin x / 5x (x -> 0) 2. lim tan^2x / x (x ->0) 3. lim cos x tan x / x (x -> 0)

**calculus **

find the limit without using L'Hopital's Rule Lim(X->-4) (16-x^2 / x+4)

**calculus again**

Suppose lim x->0 {g(x)-g(0)} / x = 1. It follows necesarily that a. g is not defined at x=0 b. the limit of g(x) as x approaches equals 1 c.g is not continuous at x=0 d.g'(0) = 1 The answer is d, can someone please explain how? Thanks. lim x->0 {g(x)-g(0)} / x = 1. You ...

**pre calc help please!!!!**

given that lim f(x) = 5 x->x and lim g(x) = -7 x -> c find lim [f(x)+g(x)]^2 x-> c a. limit does not exist b. -175 c. 4 d. 245 e. -70

**Calculus**

Show that limit as n approaches infinity of (1+x/n)^n=e^x for any x>0... Should i use the formula e= lim as x->0 (1+x)^(1/x) or e= lim as x->infinity (1+1/n)^n Am i able to substitute in x/n for x? and then say that e lim x ->0 (1+x/n)^(1/(x/n)) and then raise it ...

**Calculus**

Below are the 5 problems which I had trouble in. I can't seem to get the answer in the back of the book. Thanks for the help! lim (theta-pi/2)sec(theta) theta->pi/2 Answer: -1 I am not sure what to do here. lim (tan(theta))^(theta) theta->0+ Answer:1 ln(tan(theta))/(1/...

**Calculus**

Using l'hopitals rule, evaluate the following: Lim x-->infinity Inx/2(x)^1/2 Limx-->0 (1/x - 1/e^x -1) I tried both multiple times and I'm still not getting an answer. For the first one I tried l'hopitals rule 3 times and I'm still getting infinity over infinity for an ...

**calculus**

(lim x --> +inf) of (x^2 - [[x^2]])/2 using Squeeze theorem only thank you so much in advance! its for my midterm tonight!

**survey of clac**

g(x)={x+6, for x<-2, -1/2x+1, for x> or equal to -2, find the limit: lim g(x)= lim g(x)= x-->-2^- x-->-2^+ lim g(x)= lim g(x)= x-->-2^+ x-->4^-

**Calculus**

Find the positive integers k for which lim ->0 sin(sin(x))/x^k exists, and then find the value the limit. (hint:consider first k=0, then k=1. Find the limit in these simple cases. Next take k=2 and finally consder k>2 and find the limit in these cases as well) I did this...

**Calculus**

Find the indicated limits. If the limit does not exist, so state, or use the symbol + ∞ or - ∞. f(x) = { 2 - x if x ≤ 3 { -1 + 3x - x^2 if x > 3 a) lim 3+ f(x) x->3 b) lim 3- f(x) x->3 c) lim f(x) x->3 d) lim ∞ f(x) x->3 e) lim -∞ f...

**Calculus**

Find the indicated limits. If the limit does not exist, so state, or use the symbol + ∞ or - ∞. f(x) = { 2 - x if x ≤ 3 { -1 + 3x - x^2 if x > 3 a) lim 3+ f(x) x->3 b) lim 3- f(x) x->3 c) lim f(x) x->3 d) lim ∞ f(x) x->3 e) lim -∞ f...

**limit calc question**

2 questions!!! 1. Limit X approaching A (X^1/3-a^1/3)/ x-a 2. LiMIT x approaching 0 (1/3+x – 1/3) /x On the first, would it help to write the denominator (x-a) as the difference of two cubes ((x^1/3 cubed - a^1/3 cubed) second. use LHopitals rule. Take the derivative of the ...

**calculus-limits??**

i keep on doing something wrong on this problem,i guess im forgetting to add or subtract some number. use f'(x)=lim h-->0 f(x+h)-f(x)/h to find the limit: lim h-->0 sin^(3)4(x+h)-sin^(3)4x/h

**Calculus**

Find the following limits algebraically or explain why they don’t exist. lim x->0 sin5x/2x lim x->0 1-cosx/x lim x->7 |x-7|/x-7 lim x->7 (/x+2)-3/x-7 lim h->0 (2+h)^3-8/h lim t->0 1/t - 1/t^2+t

**Calculus**

Show that limit as n approaches infinity of (1+x/n)^n=e^x for any x>0... Should i use the formula e= lim as x->0 (1+x)^(1/x) or e= lim as x->infinity (1+1/n)^n Am i able to substitute in x/n for x? and then say that e lim x ->0 (1+x/n)^(1/(x/n)) and then raise it ...

**Calc Please Help**

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim x->1- = + infinity? lim h->0 [(-7)/(2+h^2...

**calculus**

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 (e^x − e^−x − 2x)/(x − sin(x))

**Calc 1**

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 (sin^−1 x)/7x Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 x/(tan...

**Calculus L'Hopital Rule**

lim x->0 (x)sin(x)/1-cos(x). They both go to 0 so 0/0. now take L'H and product rule of top? lim x->0 (1)sin(x)+ cos(x)(x)/1-cos(x) what next? how do i solve from here?

**Calc. Limits**

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim x->1- = + infinity? lim h->0 [(-7)/(2+h^2...

**calc**

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim x->1- = + infinity? lim h->0 [(-7)/(2+h^2...

**Math- Calculus**

I need help with these problems, I cannot find a similar example to help me in the book: 1. Find lim x->infinity (e^(-2x) + sin x). 2. Find the derivative of sqrt(9-x) using the limit process. 3. Find lim x-> -infinity (x + sqrt(x^2 + 2x)). 4. Show that the equation e^x...

**Pre-Calculus**

Which of the following shows the correct notation for “The limit of x^2 - 1 as x approaches 3. A. lim x^2-1 x->3 B. lim3 x->x^2-1 C. lim(x^2-3) x->x^2-1 D. lim(x^2-1) x->3 Thank you

**calculus help**

consider lim -->0^+ 1)sin(4x)/sin(3x) find the limit using a table of values 2)based on your answer what should the limit equal?

**Pre-Cal**

Given that lim x-> f(x)=6 and that lim x-> g(x)= -4, evaluate the following limit. Assume that c is a constant. lim -2f(x) - 12 / g(x). x->. I did -2(6) - (12/-4) -12 - (-3) -12 + 3 -9

**pre-cal**

Given that lim x->c f(x)=6 and that lim x->c g(x)= -4, evaluate the following limit. Assume that c is a constant. The x in front of the f is confusing me. lim [xf(x) + 3 g(x)]^2 x->c

**college math**

Use a table to frind the indicated limit. 3 7. lim(4x ) x-2 2 8. lim (2x + 1) x-3 x + 1 9. lim 2 x-0 x + 1

**CALCULUS LIMITS**

What is the following limit? lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) = I.) lim as n goes to infinity sigma (n and k=1) of pi/n sin(kpi/n) II.) Definite integral from 0 to pi of sin(x)dx III.) 2 A.) I only B.) II only C.) III ...

**Calculus**

What is the following limit? lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) = I.) lim as n goes to infinity sigma (n and k=1) of pi/n sin(kpi/n) II.) Definite integral from 0 to pi of sin(x)dx III.) 2 A.) I only B.) II only C.) III ...

**Calculus-Limits**

Okay, i posted this question yesterday, however, I did not really understand the answer I received. If your the one who answered my question, could you please elaborate. If not, could you try to answer this tough, for me, question. Thanks a lot. lim x-->0 Square root (1 + ...

**maths**

find the limit as x approaches 0 of the function: (e^x.sinx)/x without using hopitals rule

**Limits**

Let lim f(x) =16 as x->4 and lim g(x) =8 as x->4 Use the limit rules to find lim [cos(pi*f(x)/g(x))] as x->4

**calculus**

The limit as x approaches infinity of (e^x+x)^(1/x). I got that it diverges, but I'm not sure if I made a mistake. My work: lim(e+x^(1/x)) lim(e+(1/x^x)) lim(ex^x+1)/x^x l'hopital:lim(e^(e(lnx+1))+1)/e^(lnx+1) diverges?

**Check my CALCULUS work, please! :)**

Question 1. lim h->0(sqrt 49+h-7)/h = 14 1/14*** 0 7 -1/7 Question 2. lim x->infinity(12+x-3x^2)/(x^2-4)= -3*** -2 0 2 3 Question 3. lim x->infinity (5x^3+x^7)/(e^x)= infinity*** 0 -1 3 Question 4. Given that: x 6.8 6.9 6.99 7.01 7.1 7.2 g(x) 9.44 10.21 10.92 -11.08 -...

**MATH - Calculus (2)**

if lim f(x) = a^3 x->a and if lim g(x) = a^2 x->a calculate the following limit: lim f(x)*(x-a)/(x^3-a^3)*g(x) = x->a

**Calculus**

Evaluate each limit. If it exists. a) Lim x^3 + 1 x->1 ------- x + 1 b) Lim 3x^2 - x^3 x->0 ---------- x^3 + 4x^2 c) Lim 16 - x x->16 --------- (√x) - 4

**calculus**

calculate the lim as x-> pi (cos(x)+1)/(x-pi) using the special limit lim x->0 sinx/x

**Calculus**

Use L'Hoplital's rule to find the limit. Lim x->0 (3-3cos(x))/(sin(4x))

**calculus**

Find lim (with x¨‡ under it) (x^2-1)/x or state that the limit does not exist. I'm so lost with what to do with limits...I tried to simplify the expression to lim x+lim -1/x, but from there I'm pretty dumbfounded. If anybody could help me out I'd really appreciate it.

**Calculus**

I have two similar problems that I need help completing. Please show all your work. Question: Find the limit L. Then use the å-ä definition to prove the limit is L. 1. lim (2x+5) x->3 2. lim 3 x->6 Thank you for your anticipated help!!

**calculus**

1. integral -oo, oo [(2x)/(x^2+1)^2] dx 2. integral 0, pi/2 cot(theta) d(theta) (a) state why the integral is improper or involves improper integral (b) determine whether the integral converges or diverges converges? (c) evaluate the integral if it converges CONFUSE: how would...

**Calc-Math**

Evaluate the limit using l'hospitals rule. lim->0 (3^x-6^x)/x. Do not know where to even begin on this one... please help.

**Pre-cal**

Please determine the following limits if they exist. If the limit does not exist put DNE. lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity lim 4n-3 / 3n^2+2 n-> infinity I did lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity (2+6x-3x²)/(4x²+4x+1) (2/x² + 6/x -3)/(4 + 4/x + 1/x²) as...

**Calc 1**

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 (7x−sin 7x)/(7x−tan 7x)

**Math**

Find the limit if it exists. lim 1/(x-2) = infinity x→2+ lim 1/(x-2) = negative infinity x→2- lim 1/(x-2) = Does not exist x→2 lim (3x+2) = infinity x→∞ lim 999/(x^3) = 0 x→-∞ Can someone check to make sure that I am correct?

**Calculus..more help!**

I have a question relating to limits that I solved lim(x-->0) (1-cosx)/2x^2 I multiplied the numerator and denominator by (1+cosx) to get lim(x->0) (1-cos^2x)/2x^2(1+cosx) = lim(x->0)sin^2x/2x^2(1+cosx) the lim(x->0) (sinx/x)^2 would =1 I then substituted the ...

**math**

i need some serious help with limits in pre-calc. here are a few questions that i really do not understand. 1. Evaluate: lim (3x^3-2x^2+5) x--> -1 2. Evaluate: lim [ln(4x+1) x-->2 3. Evaluate: lim[cos(pi x/3)] x-->2 4. Evaluate: lim x^2+x-6/x^2-9 x--> -3 5. ...

**calculus**

Find the lim x->infinite (x/[3x+5]) where [] denotes the greatest integer function use the squeeze theorem.

**CALCULUS**

What is the following limit? lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) = I.) lim as n goes to infinity sigma (n and k=1) of pi/n sin(kpi/n) II.) Definite integral from 0 to pi of sin(x)dx III.) 2 A.) I only B.) II only C.) III ...

**Calculus Help Please Urgent!!!**

find the limit algebraically. Use L'Hospital's Rule where appropriate. If there is a moare elementary method, consider using it. If L'Hospital's Rule doesn't apply, explain why lim -> 0 (cot(x)-(1/x)) show work please!!!

**Calculus (lim)**

consider the statement lim x³-6x²+11x-6 / x-1 = 2 x->1 Using the definition of the limit, state what must be true for the above limit to hold, that is, for every ..., there is ..., so that.... Use a specific function and limit not just f and L. Verify the limit is true ...

**Math**

Find the limit f(x) = -(x^2)/(x+3) a) lim f(x) x-3^- b) lim f(x) x-3^+ c) lim f(x) x-3 I know that a is dne because of the graph and that b is -infinity. And that c is dne because they are not the same. But I do not know how to do this algebraically. Thank you

**calculus**

lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2) (x,y)->(1,0) my question is can you approach (1,0) with y=x and does that change the the limit to lim f(x,x) (x,x)->(1,1) in which case i get 3/4 as the limit. and if this is not how i go about doing it can you point me in the ...

**Calc 1**

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→∞ x^7e^−x^6 Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→∞...

**CALCULUS**

Evaluate each of the following. (a) lim x->0(e^x)-1-x/ x^2 (b) lim x->0 x-sinx/x^3 (c) lim x->infinity (In x)^2/x (d) lim x->0+ (sinx)In x (e) lim x->0+ (cos3x)^5/x (f) lim x->1+ ((1/x-1) -(1/In x))

**Calculus, please check my answers!**

1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 ***The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 ***2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 ***9 The limit does not exist. 4.For the ...

**Calculus - ratio test**

infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) / ((n+1)(n!)(e^n)) | ..the e^n...

**math**

I have to evaluate the limit of these problems and I can't figure them out! 1.) Lim x-> 1 x^2+x-2/x^2-x 2.) (please note that the -2 is not under the square root) Lim x->0 sqroot 4+h -2/ h

**Math(calculus)**

Evaluat d 4lowing1.lim x/|x| x-->0 2.lim x->1 sqrt(x^2+2- sqrt3)/x-1 3.lim n->~ f(n)=(1+1/n)^sqrtn 4. limx->0 f(x)= (12^x-3^x-4^x+1)/xtanx 5. Lim x->3 (x^n-3^n)^n/(n-3)^n

**Calc**

I have a test soon.. and I really need to know how to do this problem.. please help!!! lim as x-->0 sin^2(x)/tan(x^2) the answer is 1, but I have no clue how to get that! Use series expansions. Look up Taylor expansion on google and study that topic first. Using series ...

**calculus1**

Find limit if exists 1. Lim x goes to 0 x^2cos(1/x^3) 2.lim x goes to 1 to the left 4x/x^2+2x-3 3. Lim x goes to π to the right cscx

**calculus**

f(x)= 3x+2, x<0 3, x=0 x^2+1, x>0 g(x)= -x+3, x<1 0, x=1 x^2, x>1 find a) lim x->0+ g(f(x)) b)lim x->0- g(f(x)) c) lim x->1+ f(g(x)) d)lim x->1- f(g(x)) I am trying to understand these. Help appreciated. Have a test soon.

**Math**

Find the limit if it exists. lim 1/(x-2) x→2+ lim 1/(x-2) x→2- lim 1/(x-2) x→2 lim (3x+2) x→∞ lim 999/(x^3) x→-∞

**Math**

Find the limit if it exists. lim 1/(x-2) x→2+ lim 1/(x-2) x→2- lim 1/(x-2) x→2 lim (3x+2) x→∞ lim 999/(x^3) x→-∞

**Calculus help, please!**

1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 9 The limit does not exist. 4.For the function g(f)=4f...

**Calculus**

Evaluate the limit using L'Hospital's rule if necessary. lim as x goes to +infinity x^(6/x)

**Calculus**

Using L'Hopital's Rule, find the limit: lim x->0+ (as x approaches 0 from the right) of sinx/x^(1/3)

**calculus - ratio test**

infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) / ((n+1)(n!)(e^n)) | ..the e^n...

**Math**

Evaluate: (without using Hopital's rule) d) lim (6-3x)/(((5x+6)^1/2)-4) x->2

**calculus**

Using the 4 step method fidn the derivative of F(x) =1/(x^2) 1) I got /(x^2+2xh+h^2) 2)2xh+h^2/(x^2+2xh+h^2)(x^2) 3)2x+h/(x^2+2xh+h^2)(x^2) 4)1/x I don't understand why I got this problem wrong on my test can you explain the correct way? The second step is wrong, it should be ...

**Calculus**

Let f be a function defined for all real numbers. Which of the following statements must be true about f? Which might be true? Which must be false? Justify your answers. (a) lim of f(x) as x approaches a = f(a) (b) If the lim of f(x)/x as x approaches 0 = 2, then f(0)=0. (c) ...

**calculus**

It is known that x 2 + 4x ≤ f(x) ≤ -x 2 -4x the interval [-4, 0]. Use the Squeeze Theorem to evaluate the limit as x approaches negative 2 of f of x. -4 0 4 Squeeze Theorem does not apply

**Calculus**

Use pinching theorem to evaluate lim x-->1 ((x-1)sin(1/x-1)) I'm confused in the pinch theorem analytically

**Math - Calculus**

f(x) = { x^2sin(1/x), if x =/= 0 0, if x=0} a. find lim(x->0)f(x) and show that f(x) is continuous at x=0. b, find f'(0) using the definition of the derivative at x=0: f'(x)=lim(x->0) (f(x)-f(0)/x) c. Show that lim(x->0)f'(x) does not exist. In particular, this means ...

**Pre Cal help please!!**

given f(x) = { 2x^2 + 5 x < or equal to 2 3 - x^2 x > 2 find lim f(x) x -> 2^+ lim f(x) x -> 2^- lim f(x) x -> 2