# 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.

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 $$x\to0$$.

$$\lim_{x \to 0}{(\sin(x) + 2^x)^\frac{\cos x}{\sin x}}= \lim_{x \to 0}{[1+(\sin(x) + 2^x-1)]^\frac{\cos x}{\sin x}}=$$ $$=\lim_{x \to 0} \left[\left[1+(\sin(x) + 2^x-1)\right]^\frac{1}{\sin(x)+2^x-1}\right] ^{\frac{\cos x}{\sin x} (\sin(x)+2^x-1)}=$$ $$=\lim_{x \to 0} \left[\left[1+(\sin(x) + 2^x-1)\right]^\frac{1}{\sin(x)+2^x-1}\right] ^{\cos(x)\left(1+\frac{2^x-1}{\sin(x)}\right)}= e^{\lim_{x\to0}\cos(x)\left(1+\frac{2^x-1}{\sin(x)}\right)}.$$ 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 $$y\to0$$.

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