real analysis - How do I compute $lim_{x to 0}{(sin(x) + 2^x)^frac{cos x}{sin x}}$ without L'Hopital's rule?

What I've tried so far is to use the exponent and log functions: $$\lim_{x \to 0}{(\sin(x) + 2^x)^\frac{\cos x}{\sin x}}= \lim_{x \to 0}e^ {\ln {{(\sin(x) + 2^x)^\frac{\cos x}{\sin x}}}}=\lim_{x \to 0}e^ {\frac{1}{\tan x}{\ln {{(\sin(x) + 2^x)}}}}$$ .

From here I used the expansion for $\tan x$ but the denominator turned out to be zero. I also tried expanding $\sin x$ and $\cos x$ with the hope of simplifying $\frac{\cos x}{\sin x}$ to a constant term and a denominator without $x$ but I still have denominators with $x$.

Any hint on how to proceed is appreciated.

Answer

Take the logarithm and use standard first order Taylor expansions: $$\lim_{x\to0} \frac{\log\bigl(\sin(x)+2^x\bigr)}{\tan(x)} =\lim_{x\to0} \frac{\log\bigl(\sin(x)+2^x\bigr)}{x+o(x)} =\lim_{x\to0} \frac{x+\log(2)x+o(x)}{x+o(x)} = 1+\log(2).$$ Then $$\lim_{x\to0} \bigl(\sin(x)+2^x\bigr)^{\cot(x)} = e^{1+\log(2)} = 2e.$$

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EDIT

Maybe it's important to clarify why $\log\bigl(\sin(x)+2^x\bigr)=x+\log(2)x+o(x)$. I'm using the following facts:

• $\log(1+t) = t+o(t)$ as $t\to0$,


• $\sin(x)+2^x = 1+x+\log(2)x+o(x)$ as $x\to0$.