$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?

Question

please explain why my procedure is wrong i am not able to find out?? I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions). The actual answer for the given question is $\frac{1}{2}\log(2)$. My course book has shown that don't use this step but has not given the reason. AND Please TELL why i am WRONG

Answer 1

The fourth line is invalid: If $F(x)$ and $G(x)$ tend to the same limit as $x\to 0,$ we cannot conclude that $(1/x^2)(F(x)-G(x))$ tends to $0.$ For example $F(x)=4 x^2+1$ and $G(x)=x^2+1$. In general, replacing the term $1/x^2$ with any $H(x)$: If $F(x)-G(x)$ and $H(x)$ tend to $0,$ then consider $J(x)=F(x)-G(x).$ We cannot determine whether $J(x)/H(x)$ has a limit, or what the limit is if it exists, just by knowing that $J$ and $H$ tend to $0.$

Answer 2

The basic limits needed are $\frac{e^x-1}{x} \to 1 $, $\frac{\sin x}{x} \to 1 $, and $\cos x \to 1 $ as $x \to 0$.

First, as $x \to 0$,

$\begin{array}\\ \tan x -\sin x &=\frac{\sin x}{\cos x}-\sin x\\ &=\sin x(\frac{1}{\cos x}-1)\\ &=\sin x(\frac{1-\cos x}{\cos x})\\ &=\sin x(\frac{2\sin^2(x/2)}{\cos x})\\ &\to x(\frac{2(x/2)^2}{1})\\ &=x^3/2\\ \end{array} $

Then

$\begin{array}\\ \frac{2^{\tan x}-2^{\sin x}}{x^2\sin x} &=2^{\sin x}\frac{2^{\tan x-\sin x}-1}{x^2\sin x}\\ &\to \frac{2^{\tan x-\sin x}-1}{x^2\sin x} \qquad\text{since } 2^{\sin x} \to 1\\ &\to \frac{2^{x^3/2}-1}{x^3}\\ &= \frac{e^{\ln 2 x^3/2}-1}{x^3}\\ &\to \frac{\ln 2 x^3/2}{x^3}\\ &= \ln 2/2\\ \end{array} $

Answer 3

First note that the following can only be used if both resulting limits are convergent

$$\lim\limits_{x\to c}\left[f(x)\pm g(x)\right]=\left[\lim\limits_{x\to c}f(x)\right]\pm \left[\lim\limits_{x\to c}g(x)\right]$$ $$\lim\limits_{x\to c}\left[f(x)g(x)\right]=\left[\lim\limits_{x\to c}f(x)\right]\left[\lim\limits_{x\to c}g(x)\right]$$

In your first step, you separated the original limit into two divergent limits. Here are the steps in excruciating detail $$\lim\limits_{x\to 0} \frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}$$ $$=\left[\lim\limits_{x\to 0} \frac{x}{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x} - 2^{\sin x}}{x^3}\right]$$ $$=\left[\lim\limits_{x\to 0} \frac{x}{\sin x}\right]\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{x^3}\right]$$ $$=\left[\lim\limits_{x\to 0} \frac{x}{\sin x}\right]\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\tan x - \sin x}{x^3}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$ $$=\left[\lim\limits_{x\to 0} \frac{x}{\sin x}\right]\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{(\sin x)\left(\frac{\tan x}{\sin x} - 1\right)}{x^3}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$ $$=\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\sec x - 1}{x^2}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$ $$=\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\left(\tan x\right)\left(\tan\frac{x}{2}\right)}{x^2}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$ $$=\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\tan x}{x}\right]\left[\frac12\lim\limits_{x\to 0}\frac{\tan\frac{x}{2}}{\frac{x}{2}}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$ Let $s = \frac{x}{2}$, then $$\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\tan x}{x}\right]\left[\frac12\lim\limits_{s\to 0}\frac{\tan s}{s}\right]\left[\lim\limits_{x\to 0}\frac{2^{\tan x-\sin x} - 1}{\tan x - \sin x}\right]$$ Let $t = \tan x - \sin x$, then $$\left[\lim\limits_{x\to 0} 2^{\sin x}\right]\left[\lim\limits_{x\to 0}\frac{\tan x}{x}\right]\left[\frac12\lim\limits_{s\to 0}\frac{\tan s}{s}\right]\left[\lim\limits_{t\to 0}\frac{2^{t} - 1}{t}\right]$$ $$=\frac12\lim\limits_{t\to 0}\frac{2^{t} - 1}{t}$$ $$=\frac12\log 2$$

Answer 4

One can rewrite:

$$\frac{2^{\tan x}-2^{\sin x}}{x^2\sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\tan x - \sin x}{x^2 \sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{1-\cos x}{x^2 \cos x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\sin^2 x}{x^2}\frac{1}{(1+\cos x)\cos x}$$

$$=2^{\sin x}\cdot\frac{2^{\tan x-\sin x}-1}{\tan x - \sin x}\cdot \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{(1+\cos x)\cos x}\to1\cdot\ln 2\cdot 1^2\cdot \frac{1}{2\cdot1} = \frac{\ln 2}{2}$$

Answer 5

Let $u=\tan \frac x2$ and kill: \begin{align}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}&=\frac{2^{\frac{2u}{1-u^2}} - 2^{\frac{2u}{1+u^2}}}{\frac{8u}{1+u^2}\arctan^2u }\\ &=(1+u^2)2^{\frac{2u}{1+u^2}}\frac{2^{\frac{4u^3}{1-u^4}} - 1}{8u\arctan^2u }\\ \end{align} Now since $$\frac{4u^3}{1-u^4}=4\sin^3\frac x2 \times \frac{\cos \frac x2}{\cos x}$$ we have

\begin{align}\lim _{x \to 0}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x} &=\lim _{u \to 0}(1+u^2)2^{\frac{2u}{1+u^2}}\times \lim _{u \to 0}\frac{2^{\frac{4u^3}{1-u^4}} - 1}{8u\arctan^2u }\\ &=\lim _{u \to 0}\frac{2^{\frac{4u^3}{1-u^4}} - 1}{8u\arctan^2u }\\ &=\lim _{x \to 0}\frac{2^{4x^3} - 1}{8x^3 }\\ &=\frac18\times \ln 2^4 \end{align}

Answer 6

In a nutshell,

$$\frac{2^{\tan(x)}-2^{\sin(x)}}{x^2\sin(x)}\to\frac{2^{\tan(x)}-2^{\sin(x)}}{\sin^3(x)}\to\frac{2^{(s/\sqrt{1-s^2})-s}-1}{s^3}2^s\to\frac{e^{\ln(2)s^3/2}-1}{s^3}\to\frac{\ln(2)}2.$$

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• $x$ and $\sin(x)$ are interchangeable (ratio $\to1$),
• $\dfrac1{\sqrt{1-s^2}}-1=\dfrac{1-(1-s^2)}{\sqrt{1-s^2}(1+\sqrt{1-s^2})}\to\dfrac{s^2}2,$
• $\dfrac{e^{at}-1}t\to (e^{at})'|_{t=0}=a$.

Answer 7

@vim suggested Taylor expansions:

\begin{align}\lim _{x \to 0}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x} &=\lim _{x \to 0}\frac{2^{x+\frac{x^3}{3}+O(x^5)}-2^{x-\frac{x^3}{6}+O(x^5)}}{x^2(x+\frac{x^3}{3}+O(x^5))}\\ &=\lim _{x \to 0}\frac{\left(2^{x+O(x^5)}\right)\left(2^{\frac{x^3}{3}}-2^{\frac{x^3}{6}}\right)}{x^3\left(1+O(x^5)\right)}\\ &=\lim _{x \to 0}\frac{2^{x-\frac{x^3}{6}+O(x^5)}}{1+O(x^5)}\cdot\lim_{x\to0}\frac{2^{\frac{x^3}{2}} - 1}{x^3 }\\ &=1\cdot\lim_{x\to0}\frac12\cdot\frac{2^{\frac{x^3}{2}} - 1}{\frac{x^3}{2} }\\ &=\frac12\cdot\lim_{u\to0}\cdot\frac{2^{u} - 1}{u }\\ &=\frac12\cdot \ln 2 \end{align}