Question

The value of $\underset{x\to 0}{lim}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}$ is

• A. 3
• B. 4
• C. 1
• D. 2

Answer 1

Answer: (B)

4

$\underset{x\to 0}{lim}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\underset{x\to 0}{lim}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}.\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}.\frac{\sqrt{x^2+16}+4}{\sqrt{x^2+16}+4}=\underset{x\to 0}{lim}\frac{\left(x^2\right)(\sqrt{x^2+16}+4)}{\left(x^2\right)(\sqrt{x^2+1}+1)}=\underset{x\to 0}{lim}\frac{(\sqrt{x^2+16}+4)}{(\sqrt{x^2+1}+1)}=\frac{8}{2}=4$
Hence, option 'B' is correct.