# Calculate the limit help

## Homework Statement

Calculate the limit of
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}$$

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## The Attempt at a Solution

Neither multiplying with the conjugate nor trying to break out x helps me, as I'm left with "0/0" in those cases.

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lanedance
Homework Helper

do you know l'hopitals rule? could help here i think

While I do know l'Hopitals rule, we have not yet covered it in the course. The problem should be solved without using (sadly).

lanedance
Homework Helper

## Homework Statement

Calculate the limit of
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x}$$

$$\lim_{x\rightarrow \infty} \frac{5 \sqrt{x^{2}+2} + x}{2 \sqrt{x^{2}+5} +x}$$

already looking in better shape, as its not a difference term that is leading to the zero, which ws the tricky bit, so from here I'd try multiplying through by:

$$\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{x}}$$

this should change the terms containing x in the numerator & denominator from tending to infinity, to ones tending to zero...

$$\lim_{x\rightarrow \infty} \frac{5 \sqrt{x^{2}+2} + x}{2 \sqrt{x^{2}+5} +x}$$

already looking in better shape, as its not a difference term that is leading to the zero, which ws the tricky bit, so from here I'd try multiplying through by:

$$\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{x}}$$

this should change the terms containing x in the numerator & denominator from tending to infinity, to ones tending to zero...
$$\lim_{x\rightarrow \infty} \frac{5\sqrt{x^{2}+2} + x}{2\sqrt{x^{2}+5} +x} = \lim_{x\rightarrow\infty} = \frac{5}{2}\frac{x(\sqrt{1+\frac{2}{x}})+1}{x(\sqrt{1+\frac{5}{x}})+1} = \frac{5}{2}\cdot\frac{2}{2} = \frac{5}{2}$$

Cheers!

$$\lim_{x\rightarrow \infty} \frac{5\sqrt{x^{2}+2} + x}{2\sqrt{x^{2}+5} +x} =$$

$$\lim_{x\rightarrow\infty}\frac{5}{2}\cdot \lim_{x\rightarrow\infty}\frac{x\bigg( \sqrt{1+ \frac{2}{x^2}} + 1\bigg)}{x\bigg(\sqrt{1 + \frac{5}{x^2}}+1\bigg)} =$$

$$\frac{5}{2}\cdot\frac{2}{2} = \frac{5}{2}$$
There were some significant errors in this (highlighted in the

quote box) post from a user, that I felt one of the corrected

versions should be shown.

This thread is 2 years old. You've been here long enough to know not to necropost.