How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !
 A: Use L'Hospital's Rule.
If we want to avoid explicit mention of L'Hospital's Rule, rewrite our expression as
$$\frac{\sqrt{3+x}-2}{x-1}\frac{x-1}{\sqrt[3]{7+x}-2},$$
and observe that 
$$\lim_{x\to 1}\frac{\sqrt{3+x}-2}{x-1}\quad\text{and}\quad \lim_{x\to 1}\frac{\sqrt[3]{7+x}-2}{x-1}$$
 are each derivatives. 
If we want completely to avoid using a rule of differentiation, we can calculate the limits by the usual algebra tricks. 
A: You know that $$(x^3-y^3)=(x-y)(x^2+xy+y^2)$$ and $$(a^2-b^2)=(a-b)(a+b)$$ so if $a=\sqrt{3+x}$ and $b=2$, we have $(a^2-b^2)=(\sqrt{3+x}-2)(\sqrt{3+x}+2)=3+x-4$. The same calculation can be done for another identity by taking $x=\sqrt[3]{7+x}$ and $y=2$. In fact you have $$(\sqrt[3]{7+x}-2)\big((\sqrt[3]{7+x})^2+2\sqrt[3]{7+x}+2^2\big)=(\sqrt[3]{7+x})^3-2^3=7+x-8$$ Now multiply your fraction to $\frac{\sqrt{3+x}+2}{\sqrt{3+x}+2}=1$ and $\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}=1$ simultaneously, so:
 $$\frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}\times\frac{\sqrt{3+x}+2}{\sqrt{3+x}+2}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}$$ $$=\frac{3+x-4}{7+x-8}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2}$$ $$=\frac{x-1}{x-1}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2} $$ which is equal to $$\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2}$$ near $x=1$. Now, I think taking the limit when $x$ tends to $1$ is so easy. It is $3$.
A: For convenience, change variables to $t = x - 1$, so you're looking at 
$\dfrac{\sqrt{4+t}-2}{(8+t)^{1/3}-2}$.
$$ \sqrt{4+t} = 2 \sqrt{1+\frac{t}{2}} = 2 \left(1 + \frac{t}{4} + \ldots\right) $$
$$ (8+t)^{1/3} = 2 \left(1 + \frac{t}{8}\right)^{1/3} = 2 \left(1 + \frac{t}{24} + \ldots\right) $$
A: The method totally depends on how much calculus you have studied. l'hospitals rule makes life easier in most cases but in this case in this case l'hospitals rule will be tedious since to take derivatives of square roots and cuberoots is a tad bit tedious as compared to rationalisation. so i would prefer rationalisation such that we get a rational number in the denominator. also l'hospitals rule is quite sophisticated for such simple problems, and to use it is similar to differentiating a quadratic equation to find its minima and maxima instead of just completing the squares.
A: $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\lim_{x\to 1} \frac{\sqrt{3+x}-2}{x-1}\cdot \lim_{x\to 1} \frac{x-1}{\sqrt[3]{7+x}-2}$$ $$=\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}$$
Use $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$$  which gives  
$$\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}=(0.5)4^{-0.5}\cdot\frac{1}{(1/3)8^{-2/3}}=3$$
see 
