Possible L'Hospital Rule Problem?

Question

lim x->0= (e^x-1)/sin(1x)

I am not sure which rule it is or how to solve this problem.. Any help would be great!

Answer 1

Yes, you can use l'hospital's rule because when you direct subsititute you get the indeterminate form 0/0

lim x->0= (e^x-1)/sin(1x)=
lim x->0= (e^x-1)'/(sin(1x))'
lim x->0= e^x/cosx
now direct substitute
lim x->0= e^(0)/cos(0)= 1/1 = 1

Answer 2

You can use L'Hospital's rule on this one because both are continuous and differentiable functions and both are equal to 0 when x = 0. So take the derivatives (separately, not by the quotient rule) and you'll get the limit, 1.
For more info see
http://mathworld.wolfram.com/LHospitalsRule.html

Answer 3

L'well-being middle rule ability which you're taking the derivate of the perfect purely, and then take the derivate of the backside purely and notice what you get. 3. x + x^2 / a million-2x^2 a million+2x / 0-4x = a million + 2x / -4x take derivatives lower back a million is going to 0, x's develop into their constants 2/-4 -a million/2 so the cut back is -a million/2

Answer 4

Yes, when you take the limit, you get 0/0, which allows you to use l'Hopital's Rule. So take the derivative of the top and the bottom to get e^x/cos(x). Now take the limit to 0, and you'll find the answer is 1.

Answer 5

When you substitute x = 0 in the right hand side you get: 0/0
This is an indeterminate form so you can use L'Hopital's Rule.

Answer 6

perform L'Hopitals rule:

(e^x) / cos(x)

as x goes to 0, you get 1/1 = 1

so your limit equals 1