Calculating $\lim_{x\to 0} \frac{a^{\tan x} - a^{\sin x}}{\tan x - \sin x}$ without using L'Hospital rule

Question

$$\lim_{x\to 0} \frac{a^{\tan x} - a^{\sin x}}{\tan x - \sin x}$$

We weren't supposed to do this using L'Hospital's rule

So in the beginning, I added and subtracted 1 from the numerator the get into a standard limit form

$$\frac{a^x-1}{x}.$$

From then on, I got a string of standard limits but it the end, the answer just doesn't seem to match. All the time I get a $0$ and the answer is $\ln(a)$.

Answer 1

We have :\begin{aligned}\lim_{x\to 0}{\frac{a^{\tan{x}}-a^{\sin{x}}}{\tan{x}-\sin{x}}}&=\lim_{x\to 0}{a^{\sin{x}}\frac{a^{\tan{x}-\sin{x}}-1}{\tan{x}-\sin{x}}}\\ &=a^{0}\times \ln{a}\\ &=\ln{a}\end{aligned}

Because $ \lim\limits_{x\to 0}{\frac{a^{\tan{x}-\sin{x}}-1}{\tan{x}-\sin{x}}}=\lim\limits_{y\to 0}{\frac{a^{y}-1}{y}}=\ln{a} $ as you said.

Answer 2

You are going on a right track

$\begin{align} \lim_{x\to 0}\frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x} &=\lim_{x\to 0}\frac{\frac{(a^{\tan x}-1)\tan x}{\tan x}-\frac{(a^{\sin x}-1)\sin x}{\sin x}}{\tan x-\sin x}\\ &=\lim_{x\to 0}\ln a\frac{\tan x-\sin x}{\tan x-\sin x}\\ &=\ln a \end{align}$

Answer 3

$$ \frac{a^{\tan (x)} - a^{\sin (x)}}{\tan( x) - \sin (x)}= \frac{e^{\tan (x)\log(a)} - e^{\sin (x)\log(a)}}{\tan (x) - \sin (x)}\sim \frac{\Big[1+{\tan (x)\log(a)}\Big]- \Big[1+{\sin (x)\log(a)}\Big]}{\tan (x) - \sin (x) }$$ Simplify the numerator.

Answer 4

$$\lim_{x\to 0}\frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x}$$ $$=\lim_{x\to 0}\frac{e^{\tan x\ln a}-e^{\sin x\ln a}}{\tan x-\sin x}$$ $$=\lim_{x\to 0}\frac{\left(1+\frac{\tan x\ln a}{1!}+\frac{(\tan x\ln a)^2}{2!}+\ldots\right)-\left(1+\frac{\sin x\ln a}{1!}+\frac{(\sin x\ln a)^2}{2!}+\ldots\right)}{\tan x-\sin x}$$ $$=\lim_{x\to 0}\frac{\frac{(\tan x-\sin x)\ln a}{1!}+\frac{(\tan^2 x-\sin^2 x)(\ln a)^2}{2!}+\ldots}{\tan x-\sin x}$$ $$=\lim_{x\to 0}\left(\ln a+\frac{(\tan x+\sin x)(\ln a)^2}{2}+\ldots\right)$$ $$=\ln a$$