Answer by Martín-Blas Pérez Pinilla for how to solve this limit without...
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...
View ArticleAnswer by dfnu for how to solve this limit without l'hopital
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...
View ArticleAnswer by lab bhattacharjee for how to solve this limit without l'hopital
$$\lim_{x\to a}\dfrac{\sqrt[n]{x^m+a^m}-\sqrt[n]{2a^m}}{x-a}$$$$=\lim_{x\to a}\dfrac{x^m-a^m}{x-a}\cdot\dfrac1{\sum_{r=0}^{n-1}\lim_{x\to a}(x^m+a^m)^{r/n}(2a^m)^{(n-1-r)/n}}$$$$=ma^{m-1}\dfrac1{\sum...
View ArticleAnswer by TurlocTheRed for how to solve this limit without l'hopital
$u$ sub might help. Let $u=x-2$$$\lim_{u\to 0} \frac{\sqrt[3]{u^2+4u+8}-\sqrt{u+4}}{u}$$$$\frac{\sqrt[3]{u^2+4u+8}-\sqrt{u+4}}u= \frac{2\sqrt[3]{(u^2+4u)/8+1}-2\sqrt{1+u/4}}{u}\approx...
View ArticleAnswer by Stefan Lafon for how to solve this limit without l'hopital
Let $x=2+h$.$$\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...
View Articlehow to solve this limit without l'hopital
how to solve this limit without l'hopitalI have made separation of the upper fraction by adding and subtracting a suitable quantity, multiplying by the respective conjugates and nothing, some algebraic...
View Article
