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

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.

  • $\begingroup$ 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

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.


\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}


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


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.