Evaluating $\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}$ without L'Hopital I have the following limit question, where different indices of roots appear in the numerator and the denominator
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}.$$
As we not allowed to use L'Hopital, I want to learn how we can proceed algebraically.
 A: Make the change: $x-2=t^4$. Then:
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}=\lim_{t\to1}\frac{\sqrt[3]{t^4+7}-2}{t-1}=\\
\lim_{t\to1}\frac{\sqrt[3]{t^4+7}-2}{t-1}\cdot \frac{\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2}{\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2}=\\
\lim_{t\to1} \frac{(t^4+7)-2^3}{(t-1)(\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2)}=\\
\lim_{t\to1} \frac{\require{cancel} \cancel{(t-1)}(t+1)(t^2+1)}{\cancel{(t-1)}(\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2)}=\frac13.$$
A: Hint:
Numerator:
$a^3-b^3= (a-b)(a^2+ab+b^2)$.
$a=(x+5)^{1/3}$; $b=2.$
Denominator:
$c^4-d^4=(c^2-d^2)(c^2+d^2)=$
$(c-d)(c+d)(c^2+d^2);$
$c=(x-2)^{1/4};$ $d=1.$
A: A possible purely algebraic way can use 
$$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$$
Using this formula note that


*

*$\sqrt[4]{x-2} - 1 = \frac{\left(\sqrt[4]{x-2}\right)^4 - 1^4 }{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1} =  \frac{x-3 }{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1}$

*$\sqrt[3]{x+5} - 2 = \frac{\left(\sqrt[3]{x+5}\right)^3- 2^3 }{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2} = \frac{x-3 }{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2}$
So, you get
$$\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1} = \frac{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1}{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2} \stackrel{x \to 3}{\longrightarrow}\frac{4}{12} = \frac{1}{3}$$
A: Hint
To make life simpler, start using $x=y+3$ which makes
$$\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}=\frac{\sqrt[3]{y+8}-2}{\sqrt[4]{y+1}-1}$$ and use the binomial expansion or Taylor series for small values of $y$.
Just remember that
$$(y+a)^n=a^n+n  a^{n-1}y+\frac{1}{2} n(n-1)   a^{n-2}y^2+O\left(y^3\right)$$
A: It is rather unpopular to use the standard limit $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ for evaluating limits of algebraic functions, but it is my preferred approach.
Dividing the numerator and denominator of the given expression by $x-3$ and putting $x+5=u$ in numerator and $v=x-2$ in denominator we can see that the desired limit is equal to $$\dfrac{{\displaystyle \lim_{u\to 8}\dfrac{u^{1/3}-8^{1/3}} {u-8}}} {{\displaystyle \lim_{v\to 1}\dfrac{v^{1/4}-1}{v-1}}}=\dfrac{\dfrac{1}{3}\cdot 8^{-2/3}}{\dfrac{1}{4}\cdot 1^{-3/4}} =\frac{1}{3}$$
A: By definition of derivative, by $f(x)=\sqrt[3]{x+5}$ and $g(x)=\sqrt[4]{x-2}$ we have
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}=\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{x-3}\frac{x-3}{\sqrt[4]{x-2}-1}=\frac{f'(3)}{g'(3)}$$
