Wednesday, July 25, 2018

calculus - Find the value of : $lim_{xtoinfty}frac{sqrt{x-1} - sqrt{x-2}}{sqrt{x-2} - sqrt{x-3}}$



I'm trying to solve evaluate this limit



$$\lim_{x\to\infty}\frac{\sqrt{x-1} - \sqrt{x-2}}{\sqrt{x-2} - \sqrt{x-3}}.$$



I've tried to rationalize the denominator but this is what I've got



$$\lim_{x\to\infty}(\sqrt{x-1} - \sqrt{x-2})({\sqrt{x-2} + \sqrt{x-3}})$$




and I don't know how to remove these indeterminate forms $(\infty - \infty)$.



EDIT: without l'Hospital's rule (if possible).


Answer



Fill in details:



As $\;x\to\infty\;$ we can assume $\;x>0\;$ , so:



$$\frac{\sqrt{x-1}-\sqrt{x-2}}{\sqrt{x-2}-\sqrt{x-3}}=\frac{\sqrt{x-2}+\sqrt{x-3}}{\sqrt{x-1}+\sqrt{x-2}}=\frac{\sqrt{1-\frac2x}+\sqrt{1-\frac3x}}{\sqrt{1-\frac1x}+\sqrt{1-\frac2x}}\xrightarrow[x\to\infty]{}1$$




Further hint: the first step was multiplying by conjugate of both the numerator and the denominator.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...