how to solve this limit without l'hopital

Question

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

$$\lim_{x \to 2} \frac{\sqrt[3]{x^{2}+4}-\sqrt{x+2}}{x-2}$$

Answer 1

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}

Answer 2

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 $\frac 1 {12}$.

Answer 3

$u$ 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.

Answer 4

$$\lim_{x\to a}\dfrac{\sqrt[n]{x^m+a^m}-\sqrt[n]{2a^m}}{x-a}$$

$$=\lim_{x\to a}\dfrac{x^m-a^m}{x-a}\cdot\dfrac1{\sum_{r=0}^{n-1}\lim_{x\to a}(x^m+a^m)^{r/n}(2a^m)^{(n-1-r)/n}}$$

$$=ma^{m-1}\dfrac1{\sum 2a^{(n-1)m/n}} $$

Set $a=2,m=2,n=3$

Then set $a=2,m=1,n=2$

Answer 5

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...