# How to solve $\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$ without using L'Hospital Rule

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. $$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$$

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

• Where was it hiding L'Hospital's? :-) – Simply Beautiful Art Jan 23 '17 at 19:40
• 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. – Simply Beautiful Art Jan 23 '17 at 19:42
• 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 – 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 $x=(u+1)^4$:

$$\frac{\sqrt[3]{(u+1)^4+7}+2\sqrt{3(u+1)^4+1}-6}u\\=\frac{\sqrt[3]{u^4+4u^3+6u^2+4u+8}+2\sqrt{3u^4+12u^3+18u^2+12u+4}-6}u$$

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.