Evaluate the following limit: _ x 2 x^2-2 x^2+2x-4 - Vidyadip

Question

Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\text{x}^2-2}{{\text{x}^2+\sqrt{2}\text{x}-4}}$

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Answer

$\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\text{x}^2-2}{{\text{x}^2+\sqrt{2}\text{x}-4}}$ $=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\text{x}^2+2\sqrt{2}\text{x}-\sqrt{2}\text{x}-4}$ $=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\text{x}\big(\text{x}+2\sqrt{2}\big)-\sqrt{2\big(\text{x}+2\sqrt{2}\big)}}$ $=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\big(\text{x}+2\sqrt{2}\big)\big(\text{x}-\sqrt{2}\big)}$ $=\frac{\sqrt{2}+\sqrt{2}}{\sqrt{2}+2\sqrt{2}}=\frac{2\sqrt{2}}{3\sqrt{2}}$ $=\frac23$

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