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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.

E.Nole
  • 936

3 Answers3

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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 $x\to0$.
Federico
  • 5,900
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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 $y\to0$.

Tito Eliatron
  • 6,612
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In this answer, and its following comments, Bernoulli's Inequality is used to show that when $\lim\limits_{x\to a}f(x)=1$, and $\lim\limits_{x\to a}(f(x)-1)g(x)=L$, then $$ \lim\limits_{x\to a}f(x)^{g(x)}=e^L\tag1 $$ Therefore, $$ \begin{align} \lim_{x\to0}\left(\sin(x)+2^x\right)^{\cot(x)} &=\color{#C00}{\lim_{x\to0}2^{x\cot(x)}}\color{#090}{\lim_{x\to0}\left(1+\frac{\sin(x)}{2^x}\right)^{\cot(x)}}\tag{2a}\\ &=\color{#C00}{2^1}\color{#090}{e^1}\tag{2b}\\[6pt] &=2e\tag{2c} \end{align} $$ $\text{(2a)}$: the limit of a product is the product of the limits
$\text{(2b)}$: $\color{#C00}{\lim\limits_{x\to0}x\cot(x)=\lim\limits_{x\to0}\frac{x}{\tan(x)}=1}$ is proven in this answer
$\phantom{\text{(2b):}}$ and $\color{#C00}{2^x\text{ is continuous at }x=1}$
$\phantom{\text{(2b):}}$ $\color{#090}{\lim\limits_{x\to0}\frac{\sin(x)}{2^x}\cot(x)=\lim\limits_{x\to0}\frac{\cos(x)}{2^x}=1}$ and apply $(1)$
$\text{(2c)}$: simplify

robjohn
  • 345,667