Why does $\lim\limits_{x\rightarrow0}\frac{1}{2x-1} \log(2^{1+\sin{x}}-1) = 2 $?

Question

Why does $\lim\limits_{x\rightarrow0}\frac{1}{2x-1} \log(2^{1+\sin{x}}-1)= 2 $?

I have:

$\lim_{x\rightarrow0}\frac{1}{2x-1} = -1 $

$\lim_{x\rightarrow0}2^{1+\sin{x}}-1 = 1 \implies$ $\lim_{x\rightarrow0} \log(2^{1 +\sin{x}}-1) = 0$

Therefore:

$\lim_{x\rightarrow0}\frac{1}{2x-1} \log(2^{1 +\sin{x}}-1) = 0 $ ?

But I do know that it should equal to 2.

How is it so? Where have I done the mistake or how can I achieve the right result?

Answer 1

That's a typo. Most probably it should be $$\lim_{x\to 0}\frac{\log (2^{1+\sin x} - 1)}{\color{red}{2^x}-1} = 2$$

This is indeed true. L'Hospital gives for $x\to 0$ \begin{eqnarray*} \frac{\log (2^{1+\sin x} - 1)}{\color{red}{2^x}-1} & \stackrel{L'Hosp.}{\sim} & \frac{\log 2 \cdot 2^{1+\sin x} \cdot \cos x}{\log 2 \cdot 2^x \cdot (2^{1+\sin x} - 1)}\\ & \stackrel{x\to 0}{\longrightarrow} & 2 \end{eqnarray*}

Answer 2

Is the value of the limit given in the answer section and is equal to $2$? If so I believe it is a mistake. Because: $$ \lim_{x\to 0}\left(\frac{1}{2x-1}\log(2^{1+sin x} - 1)\right) \\ = \lim_{x\to 0}\left(\frac{1}{2x-1}\right)\cdot \lim_{x\to 0}\log(2^{1+\sin x} - 1) $$

Splitting the limit is valid since both limits exist. Now if you substitute $x = 0$ you may obtain: $$ \left({1\over 2\cdot 0 - 1}\right)\cdot \log(2^{1+0} - 1) = -1\cdot \log 1 = 0 $$

Therefore your limit is equal to $0$. You may also want to take a look at the graph.

Yet it's not a very formal way to find the limits it may still give you some insights. Clearly the function is crossing the origin at $x=0$ matching the results.

Answer 3

The function $f(x):=\dfrac{1}{2x-1} \log(2^{1+\sin{x}}-1), \quad x \in (- \pi/2, \pi/2)$, is continuous, hence $ \lim_{x \to 0}f(x)=f(0)=0.$

Your book is wrong ! And you are wright !