Evaluate $\lim \limits_{x \to 2} \frac{\sqrt{x+7} - \sqrt[\leftroot{5}\uproot{0}3]{4x+19}}{x-2}$

Question

I need to solve this limit, but I don't know how. I tried variable sustitutions but it didn't work out. I used $u= x+7$ or $u^3=4x+19$, for example.

$$\lim \limits_{x \to 2} \frac{\sqrt{x+7} - \sqrt[\leftroot{5}\uproot{0}3]{4x+19}}{x-2}$$

Any hint given is appreciated.

Answer 1

Assuming that you actually meant $\lim_{x\to2}$, your limit is equal to$$\lim_{x\to2}\frac{\sqrt{x+7}-3}{x-2}-\lim_{x\to2}\frac{\sqrt[3]{4x+19}-3}{x-2}.$$Can you take it from here?

Answer 2

This is equivalent to $$\begin{align} \lim_{h\to0}\frac{\sqrt{9+h}-\sqrt[3]{27+4h}}{h} &=\lim_{h\to0}\frac{3\sqrt{1+\frac{h}9}-3\sqrt[3]{1+\frac{4h}{27}}}{h}\\ &=\lim_{h\to0}\frac{3\left(1+\frac{h}{18}+o(h)\right)-3\left(1+\frac{4h}{81}+o(h)\right)}{h}\\ &=\lim_{h\to0}\frac{\frac{h}{6}-\frac{4h}{27}+o(h)}{h}\\ &=\lim_{h\to0}\left(\frac1{54}+o(1)\right)\\ &=\boxed{\frac1{54}}\\ \end{align}$$ where I have used the fact that as $x\to0$ $$(1+x)^n=1+nx+o(x)$$ otherwise known as the generalized binomial expansion.

Answer 3

Set $x-2=h$

Now as $[2,3]=6$

$$F=\dfrac{(9+h)^{1/2}-(27+4h)^{1/3}}h$$ $$=\dfrac{(9+h)^3-(27+4h)^2}h\cdot\dfrac1{\sum_{r=0}^5(9+h)^{r/2}(27+4h)^{(5-r)/3}}$$

$$=\dfrac{3\cdot9^2-216}{6\cdot3^5}=?$$