1
$\begingroup$

How do I evaluate this limit: $$\lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)$$

I tried to evaluate this using rationalizing the denominator, numerator and L'Hospital rule for nearly an hour with no success.

$\endgroup$
3
$\begingroup$

Rationalizing, observe that: \begin{align*} \lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right) &= \lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)\left(\dfrac{\sqrt{\dfrac{x^3}{x+2}} + x}{\sqrt{\dfrac{x^3}{x+2}} + x}\right) \\ &= \lim_{x\to\infty} \dfrac{\dfrac{x^3}{x+2} - x^2}{\sqrt{\dfrac{x^3}{x+2}} + x} \\ &= \lim_{x\to\infty} \dfrac{\dfrac{-2x^2}{x+2}}{\sqrt{\dfrac{x^3}{x+2}} + x} \cdot \frac{\dfrac{1}{x}}{\dfrac{1}{x}}\\ &= \dfrac{\lim\limits_{x\to\infty} \dfrac{-2x}{x+2}}{\lim\limits_{x\to\infty} \left(\sqrt{\dfrac{x}{x+2}} + 1\right)}\\ &= \frac{-2}{\sqrt{1}+1} \\ &= -1 \end{align*}

$\endgroup$
  • $\begingroup$ That's the usual approach for this kind of problem: a - b = (a - b)(a + b) / (a + b) = (a^2 - b^2) / (a + b). Tends to remove square roots and increase small differences. $\endgroup$ – gnasher729 May 18 '14 at 11:02
2
$\begingroup$

$$\sqrt{\frac{x^3}{x+2}}=x\sqrt{\frac x{x+2}}=x\sqrt{\frac1{1+2/x}}$$

$$\lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)$$

$$=\lim_{x\to\infty} x\left[\left(1+\frac2x\right)^{-\frac12}-1\right]$$

Using Taylor's Expansion, $$\lim_{x\to\infty} x\left[1-\frac12\cdot\frac2x+O\left(\frac1{x^2}\right)-1\right]$$

$\endgroup$
1
$\begingroup$

HINT :

Rewrite: $$ \frac{x^3}{x+2}=x^2-2x+4-\frac{8}{x+2} $$ and $$ \sqrt{\frac{x^3}{x+2}}-x=\left(\sqrt{\frac{x^3}{x+2}}-\sqrt{x^2}\right)\cdot\frac{\sqrt{\frac{x^3}{x+2}}+\sqrt{x^2}}{\sqrt{\frac{x^3}{x+2}}+\sqrt{x^2}}.$$

$\endgroup$
1
$\begingroup$

$\displaystyle \lim_{x \to \infty} x\cdot \sqrt{\dfrac{x}{x+2}} - x = \displaystyle \lim_{y \to 0}\dfrac{\sqrt{\dfrac{1}{1+2y}} - 1}{y} = \left(\left(1+2y\right)^{-\frac{1}{2}}\right)'|_{y=0} = -1$

$\endgroup$
0
$\begingroup$

Setting $x=\frac1h$ the limit becomes

$$\lim_{h\to0^+}\frac1h\left(\frac1{\sqrt{1+2h}}-1\right)$$

$$=\lim_{h\to0^+}\frac1h\left(\frac{1-\sqrt{1+2h}}{\sqrt{1+2h}}\right)$$

$$=\lim_{h\to0^+}\frac1h\left(\frac{1-(1+2h)}{\sqrt{1+2h}(1+\sqrt{1+2h})}\right)$$

$$=-\lim_{h\to0^+}\frac{2h}{h(1+\sqrt{1+2h})}\cdot\lim_{h\to0^+}\frac1{\sqrt{1+2h}}\cdot$$

Hope you can take it home from here

$\endgroup$
0
$\begingroup$

You can use binnomial approximation

$(1+a)^n \approx 1+ an $ $\text{ if }$ $a << 1 $

Clearly, the limit is :

$\displaystyle \lim_{x \to \infty} x \bigg(1+ \frac{2}{x}\bigg)^{-1/2} = x\bigg(1 - \dfrac{1}{x}\bigg) - x = -1$

$\endgroup$
  • $\begingroup$ Given the lateness of the Answer (about 5 months), perhaps a fuller explanation of the equivalence of the limit is in order. It certainly has quite a different appearance. $\endgroup$ – hardmath Oct 22 '14 at 18:58

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.