Question

Evaluate $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{(2+x)}^{5/2}}-{{(a+2)}^{5/2}}}{x-a}$.

Answer 1

Hint: The given problem is pretty easy and you can solve it in a few steps. Here, you can use the concept of limits and you will let a condition that when x=a, then assume the form as$\dfrac{0}{0}$. So, let’s see how we can solve the given problem.

Step-By-Step Solution:
The given problem statement is we need to evaluate$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{(2+x)}^{5/2}}-{{(a+2)}^{5/2}}}{x-a}$.
Firstly, when x=a, the expression $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{(2+x)}^{5/2}}-{{(a+2)}^{5/2}}}{x-a}$ assumes the form as$\dfrac{0}{0}$.
So, we will let$Z=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{(2+x)}^{5/2}}-{{(a+2)}^{5/2}}}{x-a}$.
If we use the formula$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{n}^{a}}-{{a}^{n}}}{n-a}=n{{a}^{n-1}}$, then also Z will not show the form$\dfrac{0}{0}$as mentioned.
So, we need to simplify, that means,
Now, we will add 2 in the denominator, but we will not change the denominator so we will subtract 2 also, that means, we get,
$\Rightarrow Z=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{(2+x)}^{5/2}}-{{(a+2)}^{5/2}}}{2+x-(a+2)}$
Now, we will let 2+x=y and a+2=k, as$x\to a;y\to k$, we get,
$\Rightarrow Z=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{y}^{5/2}}-{{k}^{5/2}}}{y-k}$
Now, we will use the formula$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{n}^{a}}-{{a}^{n}}}{n-a}=n{{a}^{n-1}}$, we get,
$\Rightarrow Z=\dfrac{5}{2}{{k}^{\dfrac{5}{2}-1}}$
$\Rightarrow Z=\dfrac{5}{2}{{k}^{\dfrac{3}{2}}}$
Now, we will place the value of k in the above equation, we get,
$\Rightarrow Z=\dfrac{5}{2}{{(a+2)}^{\dfrac{3}{2}}}$

Therefore, after evaluation of$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{(2+x)}^{5/2}}-{{(a+2)}^{5/2}}}{x-a}$, we get, $\dfrac{5}{2}{{(a+2)}^{\dfrac{3}{2}}}$.

Note:
You just need to note for the evaluation for the above question we need to check if it is in the form of $\dfrac{0}{0}$, if it is not then we will continue to simplify. Here, in this question we used the formula $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{n}^{a}}-{{a}^{n}}}{n-a}=n{{a}^{n-1}}$ in the simplification.