0
$\begingroup$

I've found the slope using $\lim_{x \to \infty } \frac{f(x)}{x}$ that is $1$ but I can't find the $h = \lim_{x \to \infty } f(x) - x$ . How we find $\lim_{x \to \infty}\frac{x\sqrt{x}}{\sqrt{x+2}} - x$ if it exists ?

Also if there is another way for finding oblique asymptote , please write it .

$\endgroup$
1
  • $\begingroup$ HINT: $\lim_{x\to\infty}\frac{x\sqrt x}{\sqrt{x+2}}-x=\lim_{x\to\infty}\frac{x(\sqrt x-\sqrt{x+2})}{\sqrt{x+2}}$ $\endgroup$ – Masacroso Dec 30 '17 at 11:23
4
$\begingroup$

$$\lim_{x\rightarrow+\infty}\left(\frac{x\sqrt{x}}{\sqrt{x+2}} - x\right)=-\lim_{x\rightarrow+\infty}\frac{2x}{(\sqrt{x+2}+\sqrt{x})\sqrt{x+2}}=$$ $$=-\lim_{x\rightarrow+\infty}\frac{2}{\left(\sqrt{1+\frac{2}{x}}+1\right)\sqrt{1+\frac{2}{x}}}=-1,$$ which gives $y=x-1$.

$\endgroup$
5
  • $\begingroup$ Why $-\lim_{x\rightarrow+\infty}\frac{2x}{(\sqrt{x+2}+\sqrt{x})\sqrt{x+2}}=-1$ ? $\endgroup$ – S.H.W Dec 30 '17 at 11:28
  • $\begingroup$ Yes, I explained it. See now. $\endgroup$ – Michael Rozenberg Dec 30 '17 at 11:29
  • $\begingroup$ I used $\frac{\sqrt{x+2}}{\sqrt{x}}=\sqrt{1+\frac{2}{x}}.$ $\endgroup$ – Michael Rozenberg Dec 30 '17 at 11:50
  • $\begingroup$ Oh , sorry . I've made a terrible mistake . Thanks for the answer . $\endgroup$ – S.H.W Dec 30 '17 at 11:51
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Dec 30 '17 at 11:52
1
$\begingroup$

$$\frac{x\sqrt{x}}{\sqrt{x+2}} - x=\frac{x\sqrt{x}-x\sqrt{x+2}}{\sqrt{x+2}}\frac{x\sqrt{x}+x\sqrt{x+2}}{x\sqrt{x}+x\sqrt{x+2}}=\frac{x^3-x^3-2x^2}{\sqrt{x+2}(x\sqrt{x}+x\sqrt{x+2})}=\frac{-2x^2}{x\sqrt{x^2+2x}+x\sqrt{x^2+4x+4}}=\frac{-2x^2}{x^2\left(\sqrt{1+\frac2x}+\sqrt{1+\frac4x+\frac4{x^2}}\right)}\to\frac{-2}{2}=-1$$

$\endgroup$
4
  • 1
    $\begingroup$ Can you explain the last step ? $\endgroup$ – S.H.W Dec 30 '17 at 11:27
  • $\begingroup$ you simply take $x^2$ from denominator, I'll add this step $\endgroup$ – user Dec 30 '17 at 11:29
  • $\begingroup$ @S.H.W is it clear now? $\endgroup$ – user Dec 30 '17 at 11:42
  • $\begingroup$ Yes , thanks a lot . $\endgroup$ – S.H.W Dec 30 '17 at 11:46
1
$\begingroup$

Alternative way. By using the Taylor expansion $(1+t)^a=1+at+o(t)$ as $t\to 0$, we can find the oblique asymptote in just one step: as $x\to +\infty$ $$f(x)=\frac{x\sqrt{x}}{\sqrt{x+2}}=x\left(1+\frac{2}{x}\right)^{-1/2}=x\left(1-\frac{1}{2}\cdot\frac{2}{x}+o(1/x)\right)=x-1+o(1).$$ Hence $\lim_{x\to +\infty}(f(x)-(x-1))=0$, which means that the oblique asymptote is $y=x-1$.

$\endgroup$
4
  • $\begingroup$ Brilliant ! Can you explain $o$ notation ? I searched over the internet but it was so confusing to understand . $\endgroup$ – S.H.W Dec 30 '17 at 11:56
  • $\begingroup$ @S.H.W This is the "little-o_notation". Take a look here: en.wikipedia.org/wiki/Big_O_notation#Little-o_notation $\endgroup$ – Robert Z Dec 30 '17 at 12:02
  • $\begingroup$ @S.H.W To avoid the "little-o_notation" just use the fact that $\lim_{t\to 0}((1+t)^a-1)/t=a$. $\endgroup$ – Robert Z Dec 30 '17 at 12:12
  • $\begingroup$ Okay , thanks a lot . $\endgroup$ – S.H.W Dec 30 '17 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.