Question

If $l=\underset{n\to 3}{lim}\frac{x^2-9}{\sqrt{x^2+7}-4}$ and $m=\underset{n\to -3}{lim}\frac{x^2-9}{\sqrt{x^2+7}-4}$, then

• A. $l\ne m$
• B. $l=2m$
• C. $l=-m$
• D. $l=m$

Answer 1

The correct option is D $l=m$

Given $l=\underset{x\to 3}{lim}\frac{x^2-9}{\sqrt{x^2+7}-4}=\frac{0}{0}$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore, $l=\underset{x\to 3}{lim}\frac{\frac{d}{dx}(x^2-9)}{\frac{d}{dx}(\sqrt{x^2+7}-4)}=\underset{x\to 3}{lim}\frac{2x}{\frac{x}{\sqrt{x^2+7}}}=\underset{x\to 3}{lim}2\sqrt{x^2+7}=2\left(\sqrt{16}\right)=8$

Given $m=\underset{x\to -3}{lim}\frac{x^2-9}{\sqrt{x^2+7}-4}=\frac{0}{0}$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore, $m=\underset{x\to -3}{lim}\frac{\frac{d}{dx}(x^2-9)}{\frac{d}{dx}(\sqrt{x^2+7}-4)}=\underset{x\to -3}{lim}\frac{2x}{\frac{x}{\sqrt{x^2+7}}}=\underset{x\to -3}{lim}2\sqrt{x^2+7}=2\left(\sqrt{16}\right)=8$

Hence, $l=8,m=8$

Therefore, $l=m$