Question

Find the limit: $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-{{e}^{-x}}-2\ln \left( 1+x \right)}{x\sin x}$.

Answer 1

Hint: Using different limit formulas we first try to break the given limit into a sum of three limits. We apply the limit theorems to find their values in number. We try to break the limit and keep the part of $\displaystyle \lim_{x \to 0}\dfrac{1}{\sin x}$ separate. The other part gives us a value of 0 and in multiplication the total value of the limit becomes 0.

Complete step-by-step answer:
To find the limit value of the given limits we are going to use some limit forms.
We know that $\displaystyle \lim_{x \to a}\left[ f\left( x \right)\pm g\left( x \right) \right]=\displaystyle \lim_{x \to a}f\left( x \right)\pm \displaystyle \lim_{x \to a}g\left( x \right)$.
Also, we have $\displaystyle \lim_{x \to a}\left[ f\left( x \right).g\left( x \right) \right]=\displaystyle \lim_{x \to a}f\left( x \right)\times \displaystyle \lim_{x \to a}g\left( x \right)$ and \[\displaystyle \lim_{x \to a}\left[ \dfrac{f\left( x \right)}{g\left( x \right)} \right]=\dfrac{\displaystyle \lim_{x \to a}f\left( x \right)}{\displaystyle \lim_{x \to a}g\left( x \right)}\].
We first try to keep only x in the denominator.
So, $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-{{e}^{-x}}-2\ln \left( 1+x \right)}{x\sin x}=\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-{{e}^{-x}}-2\ln \left( 1+x \right)}{x}\times \displaystyle \lim_{x \to 0}\dfrac{1}{\sin x}$
Let’s assume $\displaystyle \lim_{x \to 0}\dfrac{1}{\sin x}=a$ and try to solve the rest of the limit.
We have been given a number of functions and we are going to break them.
$\begin{align}
  & \displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-{{e}^{-x}}-2\ln \left( 1+x \right)}{x} \\
 & =\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-1+1-{{e}^{-x}}-2\ln \left( 1+x \right)}{x} \\
 & =\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-1}{x}+\displaystyle \lim_{x \to 0}\dfrac{1-{{e}^{-x}}}{x}-\displaystyle \lim_{x \to 0}\dfrac{2\ln \left( 1+x \right)}{x} \\
 & =\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-1}{x}-\displaystyle \lim_{x \to 0}\dfrac{{{e}^{-x}}-1}{x}-\displaystyle \lim_{x \to 0}\dfrac{2\ln \left( 1+x \right)}{x} \\
\end{align}$
We know $\displaystyle \lim_{x \to 0}\dfrac{\ln \left( 1+x \right)}{x}=1,\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-1}{x}=1$.
For the second limit in the expansion, for $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{-x}}-1}{x}$ we will change the variable to keep it true to the property.
Let’s assume $z=-x$. Now if $x \to 0\Rightarrow z\to 0$.
So, $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{-x}}-1}{x}=\displaystyle \lim_{z\to 0}\dfrac{{{e}^{z}}-1}{-z}$
So, putting the values we get
\[\begin{align}
  & \displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-1}{x}-\displaystyle \lim_{z\to 0}\dfrac{{{e}^{z}}-1}{-z}-\displaystyle \lim_{x \to 0}\dfrac{2\ln \left( 1+x \right)}{x} \\
 & =1+\displaystyle \lim_{z\to 0}\dfrac{{{e}^{z}}-1}{z}-2\displaystyle \lim_{x \to 0}\dfrac{\ln \left( 1+x \right)}{x} \\
 & =1+1-2\times 1 \\
 & =0 \\
\end{align}\]
Now we don’t need to find the value of $\displaystyle \lim_{x \to 0}\dfrac{1}{\sin x}=a$ as in the multiplication of $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-{{e}^{-x}}-2\ln \left( 1+x \right)}{x}\times \displaystyle \lim_{x \to 0}\dfrac{1}{\sin x}$. The left part value is 0. So, whatever be the answer of $\displaystyle \lim_{x \to 0}\dfrac{1}{\sin x}=a$ be the final answer will always remain 0.
So, $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-{{e}^{-x}}-2\ln \left( 1+x \right)}{x\sin x}=\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-{{e}^{-x}}-2\ln \left( 1+x \right)}{x}\times \displaystyle \lim_{x \to 0}\dfrac{1}{\sin x}=0$.
The limit value of $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{x}}-{{e}^{-x}}-2\ln \left( 1+x \right)}{x\sin x}$ is 0.

Note: We need to remember to first do the limit in rough just to understand which parts are directly solvable to find the answers. We try to form the whole limit in that order. We have tools to make them in multiplication form or summation form. $\displaystyle \lim_{x \to 0}\dfrac{1}{\sin x}$ have break in the curve at point $x=0$.