Question

How do you evaluate the limit $\dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x}$ as $x \to 0$?

Answer 1

Hint: We will directly convert the function $2{{x}^{2}}+x$ into simpler form. Also, it is important to multiply the function $\dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x}$ by $\dfrac{2x}{2x}$. This is going to be helpful here. We will use the formula \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( x \right)}{x}=1\] to solve this question further. After these processes we will substitute the value of x as 0 to get the desired answer.

Complete step-by-step answer:
In this question we need to solve the limit for the function $\dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x}$. The limit chosen here is $x \to 0$. Before solving it we will try to figure out the concept of limit in brief. A limit is simply the closeness of the given function to the given limit. Therefore, we need to find the limit of the function $\dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x}$ as x comes closer to the point 0.
To solve this question we will multiply the function $\dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x}$ by $\dfrac{2x}{2x}$. Therefore, we get
 \[\displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{2x}{2x}\times \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)\].
Now, we will change the expression into division form shown below.
\[\begin{align}
  & \displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)=\displaystyle \lim_{x \to 0}\left( \dfrac{\dfrac{\sin \left( 2x \right)}{2x}}{\dfrac{2{{x}^{2}}+x}{2x}} \right) \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)=\dfrac{\displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2x} \right)}{\displaystyle \lim_{x \to 0}\left( \dfrac{2{{x}^{2}}+x}{2x} \right)} \\
\end{align}\]
Since, \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( x \right)}{x}=1\] therefore, we can write \[\dfrac{\displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2x} \right)}{\displaystyle \lim_{x \to 0}\left( \dfrac{2{{x}^{2}}+x}{2x} \right)}=\dfrac{1}{\displaystyle \lim_{x \to 0}\left( \dfrac{2{{x}^{2}}+x}{2x} \right)}\].
\[\Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)=\dfrac{1}{\displaystyle \lim_{x \to 0}\left( \dfrac{2{{x}^{2}}+x}{2x} \right)}\]
To solve this question further we need to convert $2{{x}^{2}}+x$ into simpler form by the process of factorization. Therefore, we get $2{{x}^{2}}+x=x\left( 2x+1 \right)$. Thus,
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)=\dfrac{1}{\displaystyle \lim_{x \to 0}\left( \dfrac{x\left( 2x+1 \right)}{2x} \right)} \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)=\dfrac{1}{\dfrac{\left( 2\left( 0 \right)+1 \right)}{2}} \\
 & \Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)=\dfrac{1}{\dfrac{1}{2}}=2 \\
\end{align}\]
Hence, the correct limit of the function given to us is 2.

Note:
We have used the term $\dfrac{2x}{2x}$ and multiplied it by the function $\dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x}$. We have done this, so as to get the term \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( 2x \right)}{2x}\]. Then, it will be easier for us to use the formula \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( x \right)}{x}=1\] to solve this question further. Then, we will substitute the value of x = 0 and get the limit of the function $\dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x}$.
We cannot substitute the value of x as 0 directly. Otherwise, it will lead to no answer. Since, $\displaystyle \lim_{x \to 0}\left( \dfrac{\sin \left( 2x \right)}{2{{x}^{2}}+x} \right)=\dfrac{\sin \left( 2\left( 0 \right) \right)}{2{{\left( 0 \right)}^{2}}+0}=\dfrac{\sin \left( 0 \right)}{2{{\left( 0 \right)}^{2}}+0}=\dfrac{\sin 0}{0}$ which is undefined. So, to restrict this problem we need to simplify the function first and then substitute the value x as 0.