Respond to this Question
Similar Questions

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? 
CALCULUS
Evaluate each of the following. (a) lim x>0(e^x)1x/ x^2 (b) lim x>0 xsinx/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/x1) (1/In x)) 
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 … 
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 … 
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 … 
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? 
Math(calculus)
Evaluat d 4lowing1.lim x/x x>0 2.lim x>1 sqrt(x^2+2 sqrt3)/x1 3.lim n>~ f(n)=(1+1/n)^sqrtn 4. limx>0 f(x)= (12^x3^x4^x+1)/xtanx 5. Lim x>3 (x^n3^n)^n/(n3)^n 
Calculus
uu) lim ln(1x) as x>1 yy) lim (√(6x)2)/(√(3x)1) as x> 2 zz) lim (1(1/2)arctanx) as x>∞ bbb) lim ln1x as x>1 If these are hard to interpret with all the parentheses, you can plug them in … 
Calculus
Find the following limits algebraically or explain why they donâ€™t exist. lim x>0 sin5x/2x lim x>0 1cosx/x lim x>7 x7/x7 lim x>7 (/x+2)3/x7 lim h>0 (2+h)^38/h lim t>0 1/t  1/t^2+t 
Calculus
Find the limit. lim 5x/(x^225) x>5 Here is the work I have so far: lim 5x/(x^225) = lim 5x/(x5)(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 …