posted by .

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) + (7/4)]/h
I used a computational website to get (7/4) as the answer, but I did not get this. My work ends with: (-28)/(4(h+8) + (7(h+8)/(4(h+8)) and I end with -5.25 for an answer?

Last one!

lim x-> neg. infinity (-2x^2 + 3x - 2)/(5x^3 + 4x -x + 1)

The first two are correct.

Looking at the pattern of the third and considering the answer , I think you have a typo
The function appears to be f(x) = -7/x^2 and you are finding the derivative when x = 2

Instead of lim h->0 [(-7)/(2+h^2) + (7/4)]/h , it should be
lim h->0 [(-7)/(2+h)^2 + (7/4)]/h , notice the change in brackets.

then I get [-28 + 7(2+h)^2]/(4(2+h)^2) / h
= [ -28 + 28 + 28h + 7h^2]/(4(2+h)^2) / h
= h(28+7h)/(4(2+h)^2) / h
= (28+7h)/(4(2+h)^2
now as h ---> 0 this becomes
28/16 = 7/4

for the last one, how about dividing each term in both the numerator and denominator by the highest power of x that you see, that is , by x^3
to get the expression as

(-2/x + 3/x^2 - 2/x^3)/(5 + 4/x^2 - 1/x^2 + 1/x^3)

Now consider each term
As x becomes hugely negative, each of the terms would still be negative, but very very close to zero, so the top approaches zero, the bottom obvious approaches 5
so you have -0/5 which approaches 0 from the negative.

## Similar Questions

1. ### calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me to …
2. ### calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me to …
3. ### 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) …
4. ### Calc. Limits

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
5. ### calc

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
6. ### Calculus

Find the horizontal asymptote of f(x)=e^x - x lim x->infinity (e^x)-x= infinity when it's going towards infinity, shouldn't it equal to negative infinity, since 0-infinity = - infinity lim x-> -infinity (e^x)-x= infinity
7. ### calculus

State which of the conditions are applicable to the graph of y = f(x). (Select all that apply.) lim x→infinity f(x) = −infinity lim x→a+ f(x) = L lim x→infinity f(x) = L f is continuous on [0, a] lim x→infinity …
8. ### 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?
9. ### Math

1. If -1/infinity = infinity or -infinity ?
10. ### Math

1. If -1/infinity = infinity or -infinity ?

More Similar Questions