Question

Evaluate the following function $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}$ .

Answer 1

- Hint:For solving this question, we will use the formula $\underset{x\to t}{\mathop{\lim }}\,\dfrac{{{x}^{n}}-{{t}^{n}}}{x-t}=n{{t}^{n-1}}$ . After that, we will transform the terms in the given limit so, that we can apply the formula $\underset{x\to t}{\mathop{\lim }}\,\dfrac{{{x}^{n}}-{{t}^{n}}}{x-t}=n{{t}^{n-1}}$ directly. Then, we will solve further to get the final answer.

Complete step-by-step solution -

Given:
We have to find the value of the limit $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}$ .
Now, before we proceed we should know the following formula:
$\underset{x\to t}{\mathop{\lim }}\,\dfrac{{{x}^{n}}-{{t}^{n}}}{x-t}=n{{t}^{n-1}}...................\left( 1 \right)$
Now, we will solve the following limit:
$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}$
Now, we will write $x-a=\left( x+2 \right)-\left( a+2 \right)$ in the above limit. Then,
$\begin{align}
  & \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a} \\
 & \Rightarrow \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( x+2 \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{\left( x+2 \right)-\left( a+2 \right)}.................\left( 2 \right) \\
\end{align}$
Now, let $x+2=y$ . Then,
$\begin{align}
  & \underset{x\to a}{\mathop{\lim }}\,\left( x+2 \right) \\
 & \Rightarrow \underset{x\to a}{\mathop{\lim }}\,y=a+2 \\
\end{align}$
Now, from the above result, we conclude that as the value of $x$ approaches to $a$ , then the value of $y$ approaches to $a+2$ . So, we can write $\left( x\to a \right)=\left( y\to a+2 \right)$ and $\left( x+2 \right)=y$ in the equation (2). Then,
$\begin{align}
  & \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( x+2 \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{\left( x+2 \right)-\left( a+2 \right)} \\
 & \Rightarrow \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\underset{y\to a+2}{\mathop{\lim }}\,\dfrac{{{y}^{\dfrac{5}{2}}}-\left( a+2 \right)}{y-\left( a+2 \right)} \\
\end{align}$
Now, let $a+2=b$ so, we will write $a+2=b$ in the above equation. Then,
$\begin{align}
  & \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\underset{y\to a+2}{\mathop{\lim }}\,\dfrac{{{y}^{\dfrac{5}{2}}}-\left( a+2 \right)}{y-\left( a+2 \right)} \\
 & \Rightarrow \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\underset{y\to b}{\mathop{\lim }}\,\dfrac{{{y}^{\dfrac{5}{2}}}-b}{y-b} \\
\end{align}$
Now, we will use the formula from the equation (1) with the value of $n=\dfrac{5}{2}$ in the above equation. Then,
$\begin{align}
  & \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\underset{y\to b}{\mathop{\lim }}\,\dfrac{{{y}^{\dfrac{5}{2}}}-b}{y-b} \\
 & \Rightarrow \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\dfrac{5}{2}{{\left( b \right)}^{\left( \dfrac{5}{2}-1 \right)}} \\
 & \Rightarrow \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\dfrac{5}{2}\left( {{b}^{\dfrac{3}{2}}} \right) \\
\end{align}$
Now, as per our assumption $a+2=b$ so, we will write $b=a+2$ in the above equation. Then,
$\begin{align}
  & \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\dfrac{5}{2}\left( {{b}^{\dfrac{3}{2}}} \right) \\
 & \Rightarrow \underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\dfrac{5}{2}{{\left( a+2 \right)}^{\dfrac{3}{2}}} \\
\end{align}$
Now, from the above result, we conclude that the value of limit $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}$ will be equal to $\dfrac{5}{2}{{\left( a+2 \right)}^{\dfrac{3}{2}}}$ .
Thus, $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}=\dfrac{5}{2}{{\left( a+2 \right)}^{\dfrac{3}{2}}}$ .

Note: Here, the student should first understand the given limit $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{\left( 2+x \right)}^{\dfrac{5}{2}}}-{{\left( a+2 \right)}^{\dfrac{5}{2}}}}{x-a}$ and then proceed in the right direction to get the correct answer quickly. And we should proceed with the stepwise approach and first, we should try to transform the terms in the given limit so, that we can apply the formula $\underset{x\to t}{\mathop{\lim }}\,\dfrac{{{x}^{n}}-{{t}^{n}}}{x-t}=n{{t}^{n-1}}$ directly and then solve further without any mistake. Moreover, for the objective problem, we can easily write the result directly by analysing the terms in the given limit.