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$$\lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8}$$

I have this limit. I can't use L'Hopital. After rationalizing the numerator I get:

$$\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(x-8)(\sqrt{7+\sqrt[3]{x}}+3)}$$

Isn't there anything left to do after that? How to know if the limit just doesn't exist?

EDIT: I typed the limit wrong.

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  • $\begingroup$ I don't see what you did with the numerator (actually, I imagine it), but $$\lim_{x\to 8}\sqrt{7+\sqrt[3]{x}-3}=\lim_{x\to 8}\sqrt{4+\sqrt[3]{x}}=\sqrt{6}$$ $\endgroup$ – user228113 May 9 '15 at 20:24
  • $\begingroup$ Your approach is OK. Why don't you continue further and put $x = y^{3}$ and $y \to 2$. Then cancel $(y - 2)$ from numerator and denominator. Things will be easier after that. $\endgroup$ – Paramanand Singh May 11 '15 at 10:57
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The numerator (of the original limit) approaches 6, while the denominator approaches 0. Further, the denominator is positive as $x\to 8+$, and negative as $x\to 8-$. Hence the limit is $+\infty$ from the right and $-\infty$ from the left. Thus it doesn't exist.

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\begin{align} \lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8} &=\lim_{x \to 8} \frac{\sqrt{7+\sqrt[3]{x}}-3}{x-8} \cdot \frac{\sqrt{7+\sqrt[3]{x}}+3}{\sqrt{7+\sqrt[3]{x}}+3}\tag{1}\\ &=\lim_{x \to 8} \frac{(7+\sqrt[3]{x})-9}{(x-8)(7+\sqrt[3]{x}+3)}\tag{2}\\ &=\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(x-8)(7+\sqrt[3]{x}+3)}\tag{3}\\ &=\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(\sqrt[3]{x}^3-2^3)(7+\sqrt[3]{x}+3)}\tag{4}\\ &=\lim_{x \to 8} \frac{\sqrt[3]{x}-2}{(\sqrt[3]{x}-2)(\sqrt[3]{x}^2+2\sqrt[3]{x}+4)(7+\sqrt[3]{x}+3)}\tag{5}\\ &=\lim_{x \to 8} \frac{1}{(\sqrt[3]{x}^2+2\sqrt[3]{x}+4)(7+\sqrt[3]{x}+3)}\tag{6}\\[6pt] &=\frac{1}{(4+4+4)(7+4+3)}=\frac{1}{168}\tag{7} \end{align}

Explanation:

$(2)$: I used $(x-y)(x+y)=(x^2-y^2)$
$(5)$: I used $(x^3-y^3)=(x-y)(x^2+xy+y^2)$

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