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I tried first without L'Hôpital's rule:

$$\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}} = \frac{\sqrt{x}}{\sqrt[3]{x}} \cdot \frac{1+\frac{x}{\sqrt x}}{1+ \sqrt[3] x} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\frac{\sqrt x+x}{\sqrt x}}{\frac{\sqrt[3]{x}+x}{\sqrt[3]{x}}} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\sqrt[3]{x}(\sqrt x +x)}{\sqrt x (\sqrt[3]{x}+x)} = \frac{\sqrt x \sqrt[3]{x}(\sqrt x +x)}{\sqrt[3] x \sqrt x (\sqrt[3]x +x )} = \frac{\sqrt x +x }{\sqrt[3] x +x }$$

That didn't work, so then I tried L'Hôpital's:

$$\frac{\frac{1}{2}(x)^{-\frac{1}{2}}+1}{1+\frac{1}{3}(x)^{-\frac{1}{3}}} = \frac{\frac{1}{2\sqrt x }+1}{1+\frac{1}{{3\sqrt[3]{x}}}} = \frac{\frac{1+2\sqrt x}{2\sqrt x}}{\frac{1+3\sqrt[3]{x}}{3\sqrt[3]{x}}} = \frac{3\sqrt[3]{x}(1+2\sqrt x)}{2\sqrt x (1+3\sqrt[3]{x})} = \frac{3\sqrt[3]{x}+3\sqrt[3]{x}2\sqrt{x}}{2\sqrt{x} + 2\sqrt{x}3\sqrt[3]{x}} = ??? $$

How do I solve this? If you can solve this with and without L'Hôpital's, please do so.

3 Answers 3

9

Hint:

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

It's that simple!

  • 0
    How did you do that? – 2017-02-16
  • 0
    Divide the fraction by $x/x$. – 2017-02-16
  • 0
    Sorry to bother you, but I tried doing as you said and it didn't work. Could you exemplify in your answer? – 2017-02-16
  • 0
    $$\frac{\sqrt x+x}{x+\sqrt[3]{x}}=\frac{\frac{\sqrt x+x}x}{\frac{x+\sqrt[3]{x}}x}$$ – 2017-02-16
  • 0
    I tried doing that but I get $\frac{\frac{\sqrt x}{x}+1}{1+ \frac{\sqrt[3]{x}}{x}}$. What am I doin wrong? – 2017-02-16
  • 0
    Nothing. Just simplify the rest. – 2017-02-16
  • 0
    How? I can't seem to. – 2017-02-16
  • 1
    Can you see how $$\frac{\sqrt x}x=\frac1{\sqrt x}$$That's all I did in my answer. – 2017-02-16
  • 0
    Right, I got it now. Thanks! – 2017-02-18
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In such cases, it is easier to perform a suitable transformation to obtain integer-valued powers: this suggests letting $x = u^6$, hence we have the equivalent limit $$\lim_{x \to \infty} \frac{x^{1/2} + x}{x+x^{1/3}} = \lim_{u \to \infty} \frac{u^3 + u^6}{u^6 + u^2} = \lim_{u \to \infty} \frac{u^{-3} + 1}{1 + u^{-4}}$$ and the rest is trivial.

  • 0
    My way: I'd have chosen $$\dfrac1x=u^{\text{lcm}(2,3)}$$ – 2017-02-16
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    I've also learnt that a quicker way to calculate the limit of polinominals is to count only with the variable with the highest power. In this case $ \lim_{u \to \infty} \frac{u^3 + u^6}{u^6 + u^2} = \lim_{u \to \infty} \frac{u^6}{u^6} = 1 $ – 2017-02-16
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    @MarkRead note that basically is proven by the methods here. – 2017-02-16
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set ${ t }^{ 6 }=x$ then we have $$\lim _{ x\rightarrow +\infty } \frac { \sqrt { x } +x }{ x+\sqrt [ 3 ]{ x } } =\lim _{ t\rightarrow +\infty } \frac { { t }^{ 3 }+{ t }^{ 6 } }{ { t }^{ 6 }+{ t }^{ 2 } } =\lim _{ t\rightarrow +\infty } \frac { { \left( \frac { 1 }{ { t }^{ 3 } } +1 \right) } }{ \left( \frac { 1 }{ { t }^{ 4 } } +1 \right) } =1$$