Limits and Derivatives — MATHS STD 11 Science — Question

3. $1$

Solution:

Given $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ 

$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\big(\sqrt{\text{x}+1}-\sqrt{1-\text{x}}\big)\big(\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big)}$ 

$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\text{x}+1-1+\text{x}}$

$=\frac{1}{2}\cdot\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\big[\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big]$

Taking limit, we get

$=\frac{1}{2}\times1\times\big[\sqrt{0+1}+\sqrt{0-1}\big]=\frac{1}{2}\times1\times2$

$=1$