Evaluate the following limit: _ x 0 x^2+1- x - Vidyadip

Question

Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2+1-\cos\text{x}}{\text{x}\sin\text{x}}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2+1-\cos\text{x}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^2+2\sin^2\frac{\text{x}}{2}}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^2\Bigg[1+2\bigg(\frac{\sin\frac{\text{x}}{2}}{\text{x}}\bigg)^2\Bigg]}{\text{x}\sin\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1+2\Bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)^2\times\frac{1}{4}}{\frac{\sin\text{x}}{\text{x}}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{1+\lim\limits_{\text{x} \rightarrow0}2\Bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)^2\times\frac{1}{4}}{\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}}$ $\frac{1+2\times1\times\frac{1}{4}}{1}=\frac{1+\frac{1}{2}}{1}$ $=\frac{3}{2}$

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