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What is the value of the limit f(x) = \(\frac{x^2+\sqrt {2x}}{x^2-4x}\) if x approaches infinity?(a) 0(b) 2(c) 1/2(d) 4I got this question during an online interview.My enquiry is from Limits in portion Limits and Derivatives of Mathematics – Class 11

Answer»

The correct choice is (a) 0

To explain I would SAY: This is of the FORM \(\frac{\infty}{\infty}\), therefore we use L’Hospital’s rule and differentiate the numerator and DENOMINATOR.

= \(\lim\limits_{x \rightarrow \infty}\frac{2x+\sqrt{2/x}}{2x–4}\)

= \(\lim\limits_{x \rightarrow \infty}⁡\)√2 x^-3/2

= 0



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