The value of limx→0 x2+1−1×2+16−4 is

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

A
3
B
4
C
1
D
2

Solution:

$\frac{\sqrt{x^{2} + 1} - 1}{\sqrt{x^{2} + 16} - 4} = \frac{x^{2} (\sqrt{x^{2} + 16} + 14)}{x^{2} (\sqrt{x^{2} + 1} + 1)},$ for $x \neq 0$

So the required limit $= \frac{4 + 4}{1 + 1} = 4$ .

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