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\( \lim _{x \rightarrow \infty} \frac{x}{\sqrt{4 x^{2}+1}-1} \) |
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Answer» \(\lim\limits_{x\to \infty}{\frac{x}{\sqrt{4x^2+1-1}}}\) = \(\lim\limits_{x\to \infty}\cfrac1{\sqrt{4+\frac1{x^2}-\frac1x}}\) = \(\cfrac1{\sqrt{4+\frac1{\infty}}-\frac1{\infty}}\) = \(\frac1{\sqrt4}\) = \(\frac12\) (\(\because\frac1{\infty}=0\)) |
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