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.
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6 Answers
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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*}
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$\begingroup$ 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. $\endgroup$ May 18, 2014 at 11:02
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$$\sqrt{\frac{x^3}{x+2}}=x\sqrt{\frac x{x+2}}=x\sqrt{\frac1{1+2/x}}$$
$$\lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)$$
$$=\lim_{x\to\infty} x\left[\left(1+\frac2x\right)^{-\frac12}-1\right]$$
Using Taylor's Expansion, $$\lim_{x\to\infty} x\left[1-\frac12\cdot\frac2x+O\left(\frac1{x^2}\right)-1\right]$$
answered May 18, 2014 at 7:36
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$\displaystyle \lim_{x \to \infty} x\cdot \sqrt{\dfrac{x}{x+2}} - x = \displaystyle \lim_{y \to 0}\dfrac{\sqrt{\dfrac{1}{1+2y}} - 1}{y} = \left(\left(1+2y\right)^{-\frac{1}{2}}\right)'|_{y=0} = -1$
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HINT :
Rewrite: $$ \frac{x^3}{x+2}=x^2-2x+4-\frac{8}{x+2} $$ and $$ \sqrt{\frac{x^3}{x+2}}-x=\left(\sqrt{\frac{x^3}{x+2}}-\sqrt{x^2}\right)\cdot\frac{\sqrt{\frac{x^3}{x+2}}+\sqrt{x^2}}{\sqrt{\frac{x^3}{x+2}}+\sqrt{x^2}}.$$
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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
answered May 18, 2014 at 7:42
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You can use binnomial approximation
$(1+a)^n \approx 1+ an $ $\text{ if }$ $a << 1 $
Clearly, the limit is :
$\displaystyle \lim_{x \to \infty} x \bigg(1+ \frac{2}{x}\bigg)^{-1/2} = x\bigg(1 - \dfrac{1}{x}\bigg) - x = -1$
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1$\begingroup$ 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. $\endgroup$– hardmathOct 22, 2014 at 18:58