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Evaluate \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x}{\sqrt{1+2x^2}}\) |
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Answer» Correct Answer - Option 3 : \(\frac{1}{\sqrt 2}\) Calculation: We have to find the value of \(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\) \(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\) [Form \(\frac{∞}{∞}\)] This limit is of the form \(\frac{∞}{∞}\), Here, We can cancel a factor going to ∞ out of the numerator and denominator. \(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\) = \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x}{x\sqrt{\frac{1}{x^2}+2}}\) Factor x becomes ∞ at x tends to ∞, So we need to cancel this factor from numerator and denominator. = \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1}{x^2}+2}}\) = \(\frac{1}{\sqrt{\frac{1}{\infty^2}+2}}=\frac{1}{\sqrt{0+2}}=\frac{1}{\sqrt 2}\) |
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