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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.

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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 $$ \lim_{x\to0} \bigl(\sin(x)+2^x\bigr)^{\cot(x)} = e^{1+\log(2)} = 2e. $$


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 $x\to0$.
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$$\lim_{x \to 0}{(\sin(x) + 2^x)^\frac{\cos x}{\sin x}}= \lim_{x \to 0}{[1+(\sin(x) + 2^x-1)]^\frac{\cos x}{\sin x}}=$$ $$=\lim_{x \to 0} \left[\left[1+(\sin(x) + 2^x-1)\right]^\frac{1}{\sin(x)+2^x-1}\right] ^{\frac{\cos x}{\sin x} (\sin(x)+2^x-1)}=$$ $$=\lim_{x \to 0} \left[\left[1+(\sin(x) + 2^x-1)\right]^\frac{1}{\sin(x)+2^x-1}\right] ^{\cos(x)\left(1+\frac{2^x-1}{\sin(x)}\right)}= e^{\lim_{x\to0}\cos(x)\left(1+\frac{2^x-1}{\sin(x)}\right)}. $$ But $$\lim_{x\to0}\frac{2^x-1}{\sin x} = \lim_{x\to0} \frac{e^{x\log2}-1}{\sin x}=\lim_{x\to0}\frac{x\log2}{x}=\log2. $$ So your limit is equal to $e^{1+\log2}=2e$.

PD: We use that $e^{y}-1\sim y$ and $\sin y \sim y$ when $y\to0$.

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  • $\begingroup$ How is $${\frac{\cos x}{\sin x} (\sin(x)+2^x-1)}={\cos(x)+\frac{2^x-1}{\sin(x)}}?$$ $\endgroup$ – E.Nole Dec 20 '18 at 19:14
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    $\begingroup$ sorry, the last fraction must be multiplied by $\cos x$. But this does not affect thhe final result $\endgroup$ – Tito Eliatron Dec 20 '18 at 19:15
  • $\begingroup$ Now it seems correct. $\endgroup$ – Tito Eliatron Dec 20 '18 at 19:17

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