Compute the following limit. I've tried using l'Hospital. And it work the result was $\dfrac{7}{3}$. But how can I do this without this rule? I am trying to help I friend who hasn't done derivatives yet. $$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$$

  • $\begingroup$ i can not see where the square root ends and where it begins $\endgroup$ – Dr. Sonnhard Graubner Jan 23 '17 at 19:21
  • $\begingroup$ it begins at $3x$ and it ends at $1$ so $3x+1$ is inside the square root and the numerator indeed becomes zero for $x=1$ $\endgroup$ – imranfat Jan 23 '17 at 19:22
  • $\begingroup$ Your L'Hospital's limit was done incorrectly. The correct limit is $\frac{19}3$ btw $\endgroup$ – Simply Beautiful Art Jan 23 '17 at 19:35
  • $\begingroup$ desmos.com/calculator/dekc2c7k4i $\endgroup$ – Simply Beautiful Art Jan 23 '17 at 19:37

Letting $u=\sqrt[4]{x}$, you have that: $\frac{\sqrt[4]x -1}{x-1}=\frac{u-1}{u^4-1}=\frac{1}{1+u+u^2+u^3}\to \frac{1}{4}$ as $u\to 1$ and hence also as $x\to 1$.

Letting $v=\sqrt[3]{x+7}$ we get that $\frac{\sqrt[3]{x+7}-2}{x-1}=\frac{v-2}{v^3-8}=\frac{1}{v^2+2v+4}\to \frac{1}{12}$ as $v\to 2$ and hence as $x\to 1$.

Finally, let $w=\sqrt{3x+1}$ we get that $\frac{2\sqrt{3x+1}-4}{x-1}=2\frac{w-2}{\frac{1}{3}(w^2-4)}=6\frac{1}{w+2}\to \frac{3}{2}$ as $w\to 2$ and hence as $x\to 1$.

So the limit is $$\frac{\frac{1}{12}+\frac{3}{2}}{\frac{1}{4}}=\frac{19}{3}.$$

This is, of course, just hiding L'Hopital.

  • $\begingroup$ Where was it hiding L'Hospital's? :-) $\endgroup$ – Simply Beautiful Art Jan 23 '17 at 19:40
  • $\begingroup$ Ah, I see. You multiplied everything by $$\frac{\frac1{x-1}}{\frac1{x-1}}$$but I still don't see why you would call it hiding L'Hospital's rule. $\endgroup$ – Simply Beautiful Art Jan 23 '17 at 19:42
  • $\begingroup$ I'm rewriting $\frac{f(x)}{g(x)}$ with $f(x)\to 0$ and $g(x)\to 0$ as $x\to 1$ as: $$\frac{\frac{f(x)-f(1)}{x-1}}{\frac{g(x)-g(1)}{x-1}}$$ and then computing the derivatives $f'(1)$ and $g'(1)$ without ever calling it a derivative. @SimplyBeautifulArt $\endgroup$ – Thomas Andrews Jan 23 '17 at 19:56

HINT: Your expression is $$ \frac{\sqrt[3]{x+7}-2}{\sqrt[4]{x}-1}+2\frac{\sqrt{3x+1}-2}{\sqrt[4]{x}-1} $$ Then use the expression $(a^n-b^n)=(a-b)(......)$ for $n=2$, $3$, $4$.


Hint: $$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}=\\ \lim_{x\to1}\dfrac{\sqrt[3]{x+7}-2}{\sqrt[4]{x}-1}+\lim_{x\to1}\dfrac{2\sqrt{3x+1}-4}{\sqrt[4]{x}-1}\\ \lim_{x\to1}\dfrac{\sqrt[3]{x+7}-2}{\sqrt[4]{x}-1}\times\dfrac{(\sqrt[3]{x+7}+2)(\sqrt[4]{x}+1)}{(\sqrt[3]{x+7}+2)(\sqrt[4]{x}+1)}\dfrac{\sqrt{x}+1}{\sqrt{x}+1}\\+\lim_{x\to1}\dfrac{2(\sqrt{3x+1}-2)}{\sqrt[4]{x}-1}\times \dfrac{(\sqrt{3x+1}+2)(\sqrt[4]{x}+1)}{(\sqrt{3x+1}-2)(\sqrt[4]{x}+1)}\times\dfrac{\sqrt{x}+1}{\sqrt{x}+1}\\$$


Let $x=(u+1)^4$:


then some binomial expansions:

$$\sqrt[3]{8+(u^4+4u^3+6u^2+4u)}=2+\frac1{12}(u^4+4u^3+6u^2+4u)+\mathcal O(u^2)$$

$$\sqrt{4+(3u^4+12u^3+18u^2+12u)}=2+\frac14(3u^4+12u^3+18u^2+12u)+\mathcal O(u^2)$$

Thus, the limit reduces down to

$$\frac{2+\frac1{12}(u^4+4u^3+6u^2+4u)+4+\frac12(3u^4+12u^3+18u^2+12u)-6+\mathcal O(u^2)}u\\=\frac1{12}(u^3+4u^2+6u+4)+\frac12(3u^3+12u^2+18u+12)+\mathcal O(u)\\\to\frac1{12}(4)+\frac12(12)=\frac{19}3$$

which is the correct limit.


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.