Question

If $L=\underset{x\to 0}{lim}\frac{a-\sqrt{a^2-x^2}-{\displaystyle \frac{x^2}{4}}}{x^4}$, $a>0$. If L is finite, then:

• A. $a=2$
• B. $a=1$
• C. $a=\frac{1}{3}$
• D. None of these

Answer 1

The correct option is A

$a=2$

Explanation for the correct option.

Step 1: Apply L' Hospital Rule.

$\underset{x\to 0}{lim}\frac{a-\sqrt{a^2-x^2}-{\displaystyle \frac{x^2}{4}}}{x^4}$is in $\frac{0}{0}$form, so we will apply L'Hospital Rule, according to which $\underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$. So,

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{a-\sqrt{a^2-x^2}-{\displaystyle \frac{x^2}{4}}}{x^4}& =& \underset{x\to 0}{lim}\frac{-{\displaystyle \frac{1\times -2x}{2\sqrt{a^2-x^2}}}-{\displaystyle \frac{1}{4}\left(2x\right)}}{4x^3}\\ & =& \underset{x\to 0}{lim}\frac{{\displaystyle \frac{x}{\sqrt{a^2-x^2}}}-{\displaystyle \frac{x}{2}}}{4x^3}\\ & =& \underset{x\to 0}{lim}\frac{{\displaystyle \frac{1}{\sqrt{a^2-x^2}}-\frac{1}{2}}}{4x^2}\\ & =& \underset{x\to 0}{lim}\frac{{\displaystyle \frac{2-\sqrt{a^2-x^2}}{2\sqrt{a^2-x^2}}}}{4x^2}\\ & =& \underset{x\to 0}{lim}\frac{2-\sqrt{a^2-x^2}}{8x^2\sqrt{a^2-x^2}}\end{array}$

Step 2: Find the value of $a$.

By applying limit we get

$\underset{x\to 0}{lim}\frac{2-\sqrt{a^2-x^2}}{8x^2\sqrt{a^2-x^2}}=\frac{2-a}{0}$

Now as $L$is finite, so

$\begin{array}{rcl}2-a& =& 0\\ \Rightarrow a& =& 2\end{array}$

Hence, option A is correct.