Calculus - Limits Involving Trigonometry

Question

How would I go about solving this with tanx being an exponential?

lim (sin x)^{tan x}

x-->0⁺

Answer 1

Hello Russell,

When x approaches 0, (sin x)^{tan x} approaches 0⁰ which is called an indeterminate form. This can be converted into 0/0 form by using a logarithm and some algebra.

Start by letting y = (sin x)^{tan x}.

Then ln y = ln (sin x)^{tan x}.

Then using a logarithm property, ln y = tan x ln (sin x).

Now as x approaches 0⁺, the right side approaches 0 (-∞). By putting one of the two factors in the denominator, you can make a fraction. If you move the tan x to the denominator as 1/tan x or cot x, the fraction will approach -∞/∞. Then you can use L'Hopital's Rule to finish the problem and find the limit of ln y. After that, use this answer to find lim y.

Now see if you can write out all the steps and finish solving the problem.

Answer 2

I am not sure how to justify it analytically, but if you graph this function which is continuous for x>0, it certainly appears that the limit is 1. At least as far as I can see l'Hospital's rule doesn't help immediately.