I'm practicing my skills on limits and I seem to get an incorrect answer for this problem. Here's the problem

$\lim_{x\to \infty} (5x\sqrt{x^2+2}-5x^2)$.

And here's my attempt enter image description here

$$\lim_{x\to \infty} (5x^2\sqrt{1+2/x^2}-5x^2)=\lim_{x\to \infty}5x^2(\sqrt{1+2/x^2}-1)=0$$

Instead of writing a solution of yours, please indicate the mistakes in my calculations for I seem to unable to spot a mistake. Thanks.

I'd be glad if someone reformatted the problem too, can't remember the format to enter expressions here.

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Clement C. Aug 18 '16 at 18:22
  • $\begingroup$ I would have done so right away but i dont know how to format here properly and i thought it would be worse to read it like that. $\endgroup$ – Συλχαν Aug 18 '16 at 18:23
  • $\begingroup$ How did you directly conclude that the limit is 0? $\endgroup$ – Tushant Mittal Aug 18 '16 at 18:24
  • $\begingroup$ 2/x^2 when x goes to infinity is 0. Sqrt(1-0) is 1. 1-1=0 $\endgroup$ – Συλχαν Aug 18 '16 at 18:25
  • $\begingroup$ This part is correct, but the conclusion you get overall is wrong. You have "$\infty\cdot 0$", this is indeterminate... recall that you multiply by $5x^2$. $\endgroup$ – Clement C. Aug 18 '16 at 18:29

Your mistake is when you write: $$ \lim_{x\to\infty} 5x^2\left(\sqrt{1+\frac{2}{x^2}}-1\right) = 0. $$ This is not true. The inside of the parenthesis goes to $0$, indeed; but it's multiplied by $5x^2$, that blows up to $\infty$. Hence an "$\infty\cdot 0$" indeterminate form, preventing you from concluding so fast.

  • If you are familiar with Taylor expansions or asymptotic equivalents, this is obvious from the fact that (as $u\stackrel{\rm def}{=}\frac{2}{x^2}\xrightarrow[x\to\infty]{} 0$) $$ \sqrt{1+u} - 1 \operatorname*{\sim}_{u\to 0} \frac{u}{2} $$ i.e. "the parenthesis goes to $0$ at the same 'speed' as $\frac{u}{2}$." But the parenthesis is multiplied by $5x^2 = \frac{10}{u}$...

  • If you are not familiar with this, rewrite it as a derivative. $$ \frac{\sqrt{1+u}-f(0)}{u} = f'(0) $$ where $f$ is defined by $f(x) = \sqrt{1+x}$. Now, $$ 5x^2\left(\sqrt{1+\frac{2}{x^2}}-1\right) = 10\frac{f(\frac{2}{x^2})-f(0)}{\frac{2}{x^2}}. $$ and now, you can conclude. (This "make a derivative appear" is a standard trick, it gets easier with practice. I'd highly recommend learning about asymptotic equivalents (Landau notation, etc.) and Taylor expansions as soon as you can, though. It is systematic, powerful, versatile.)

  • $\begingroup$ Thanks for this explanation. I havent learned mentioned theories but i will do as soon as im able to $\endgroup$ – Συλχαν Aug 18 '16 at 18:34
  • $\begingroup$ In a nutshell: Taylor appaoximations are a way to locally approximate a function by a low-degree polynomial (typically, an affine function). So for instance, when $x\to 0$ we have $e^x = 1+ x + o(x)$, meaning "$e^x$ locally behaves like $1+x$ when sufficiently close to $0$." $\endgroup$ – Clement C. Aug 18 '16 at 18:36
  • $\begingroup$ Aha, i understand it. Thanks one more time. $\endgroup$ – Συλχαν Aug 18 '16 at 18:38

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