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