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

Question

Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big\{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big\}$

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Answer

$\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big\{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big\}$ $=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big[\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big]\times\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}\big)}{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2+1}\big)}$ $=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\frac{\text{x}\big(\text{x}^2+1-\text{x}^2+1\big)}{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}(2)}{\text{x}\Big(\sqrt{1+\frac{1}{\text{x}^2}}+\sqrt{1-\frac{1}{\text{x}^2}}\Big)}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{2}{\sqrt{1+\frac{1}{\text{x}^2}}+\sqrt{1-\frac{1}{\text{x}^2}}}$ $=\frac{2}{2}=1$ $=1$

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