how to solve this limit without l'hopital how to solve this limit without l'hopital
I have made separation of the upper fraction by adding and subtracting a suitable quantity, multiplying by the respective conjugates and nothing, some algebraic trick.
the result with l'hopital is 1/16
$$\lim_{x \to 2}  \frac{\sqrt[3]{x^{2}+4}-\sqrt{x+2}}{x-2}$$
 A: Let $x=2+h$.
$$\begin{split}
\frac{\sqrt[3]{x^{2}+4}-\sqrt{x+2}}{x-2} &= \frac{\sqrt[3]{8+4h+h^2}-\sqrt{h+4}}{h}\\
&=\frac{2\left(1+\frac{h}{2}+\frac{h^2}{8}\right)^{\frac 1 3}-2\left(1+\frac h 4\right)^{\frac 1 2}}{h}\\
&=\frac{2\left(1+\frac{h}{6}+\mathcal O(h^2)\right)-2\left(1+\frac h 8 +\mathcal O(h^2)\right)}{h}\\
&=\frac{\frac{h}{12}+\mathcal O(h^2)}{h}
\end{split}$$
Thus the limit is $\frac 1 {12}$.
A: $$\lim_{x\to a}\dfrac{\sqrt[n]{x^m+a^m}-\sqrt[n]{2a^m}}{x-a}$$
$$=\lim_{x\to a}\dfrac{x^m-a^m}{x-a}\cdot\dfrac1{\sum_{r=0}^{n-1}\lim_{x\to a}(x^m+a^m)^{r/n}(2a^m)^{(n-1-r)/n}}$$
$$=ma^{m-1}\dfrac1{\sum 2a^{(n-1)m/n}} $$
Set $a=2,m=2,n=3$
Then set $a=2,m=1,n=2$
A: $u$ sub might help. Let $u=x-2$
$$\lim_{u\to 0} \frac{\sqrt[3]{u^2+4u+8}-\sqrt{u+4}}{u}$$
$$\frac{\sqrt[3]{u^2+4u+8}-\sqrt{u+4}}u= \frac{2\sqrt[3]{(u^2+4u)/8+1}-2\sqrt{1+u/4}}{u}\approx \frac{2[1+\frac{u^2+4u}{24}]-2[1+\frac{u}{8}]}{u}$$
$$\frac{u+4}{12}-\frac{3}{12}=\frac{1}{12}$$
That about does it. You might want to prove the bounds on $(1+x)^{1/n}\approx 1+x/n$ for small x.
A: Another path using the fundamental limit
$$\frac{(1+\alpha)^k-1}{\alpha} \to k, \ \ \mbox{when} \ \ \alpha\to 0.$$
Here is my proposal.
\begin{eqnarray}
\mathcal L &=& \lim_{x\to 2}\frac{\sqrt[3]{x^2+4}-\sqrt{x+2}}{x-2}=\\
&=&\lim_{t\to 0}\frac{\sqrt[3]{t^2+4t+8}-\sqrt{t+4}}t=\\
&=& \lim_{t\to 0}\frac{2\sqrt[3]{1+\frac{t^2+4t}8}-2\sqrt{1+\frac{t}4}}{t}=\\
&=&2\lim_{t\to 0}\left(\frac{\sqrt[3]{1+\frac{t^2+4t}8}-1}{t}-\frac{\sqrt{1+\frac{t}4}-1}{t}\right)=\\
&=&2 \lim_{t\to 0}\left(\frac13 \frac{t^2+4t}{8t}-\frac12\frac{t}{4t}\right)=\\
&=&2\left(\frac16-\frac18\right)=\frac1{12}.
\end{eqnarray}
A: Dirty trick:
$$
\lim_{x\to 2}\frac{\sqrt[3]{x^2 + 4} - \sqrt{x + 2}}{x - 2} = 
\lim_{x\to 2}\frac{\sqrt[3]{x^2 + 4} - 2}{x - 2} - 
\lim_{x\to 2}\frac{\sqrt{x + 2} - 2}{x - 2}.
$$
Now, each limit is an incremental quotient (derivative). If this counts as "without l'Hôpital" is debatable...
