Question

Having trouble getting started with this one:
Evaluate the following limit
lim [(e^tanx)-1]/2x
x->0

Answer 1

Have you learned L'Hopital's rule yet?

Answer 2

First, look at the individual limits of the numerator and the denominator. Since \(e^u\) is a continuous function of u, you can take the limit through the function as follows:\[\huge \lim_{x \rightarrow 0}\left [ e^{\tan (x)}\right ] -1=e^{\underset{x \rightarrow 0}{\lim} \tan (x)}-1=0\]Do you understand?

Answer 3

And of course the limit of the denominator is also zero. You have an indeterminate form, and two differentiable functions, so you can apply L'Hopitals rule.
\[\frac{d}{dx}\left [ e^{\tan (x)}-1\right ] =\sec^2 (x) e^{\tan (x)}\]\[\frac{d}{dx}2x=2\]\[\huge \lim_{x \rightarrow 0} \frac{e^{\tan (x)}-1}{2x}=\lim_{x \rightarrow 0} \frac{\sec^2 (x)e^{\tan (x)}}{2}\]\[=\frac{(1)(1)}{2}=0.5\]

Answer 4

thank you this really helped me