A gain and A gain.

Calculus Level 3

Daniel is learning the techniques to evaluate \frac{\infty}{\infty} and 0 0 \frac{0}{0} , types of limit problems. One of the technique that he recently learned is L'Hôpital's rule . lim x x x 2 1 \lim_{x\to \infty}\dfrac{x}{\sqrt {x^2-1}}

After solving some of the problems using only this technique. He tries to solve the above limit problem.

Will he be able to solve the problem above only using the technique that he learnt?

No Yes

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1 solution

Chew-Seong Cheong
Nov 13, 2018

No , because the process is recursive as follows:

lim x x x 2 1 = lim x L’H o ˆ pital’s x 2 1 x = lim x L’H o ˆ pital’s x x 2 1 = lim x L’H o ˆ pital’s x 2 1 x = \lim_{x \to \infty} \frac x{\sqrt{x^2-1}} = \lim \limits_{x \to \infty}^{\text{L'Hôpital's}} \frac {\sqrt{x^2-1}}x = \lim \limits_{x \to \infty}^{\text{L'Hôpital's}} \frac x{\sqrt{x^2-1}} = \lim \limits_{x \to \infty}^{\text{L'Hôpital's}} \frac {\sqrt{x^2-1}}x = \cdots

The limit can be easily solved as lim x x x 2 1 = lim x 1 1 1 x 2 = 1 \displaystyle \lim_{x \to \infty} \frac x{\sqrt{x^2-1}} = \lim_{x \to \infty} \frac 1{\sqrt{1-\frac 1{x^2}}} = 1 .

Exactly!! I find this problem interesting and L-Hôpital's rule fails. Sir, are there other such functions ?

Naren Bhandari - 2 years, 7 months ago

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Read this section .

Pi Han Goh - 2 years, 7 months ago

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Thank you Sir , @Pi Han Goh .

Naren Bhandari - 2 years, 6 months ago

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