# 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.

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 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 $$\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1} = \frac{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1}{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2} \stackrel{x \to 3}{\longrightarrow}\frac{4}{12} = \frac{1}{3}$$

Hint:

Numerator:

$$a^3-b^3= (a-b)(a^2+ab+b^2)$$.

$$a=(x+5)^{1/3}$$; $$b=2.$$

Denominator:

$$c^4-d^4=(c^2-d^2)(c^2+d^2)=$$

$$(c-d)(c+d)(c^2+d^2);$$

$$c=(x-2)^{1/4};$$ $$d=1.$$

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 $$(y+a)^n=a^n+n a^{n-1}y+\frac{1}{2} n(n-1) a^{n-2}y^2+O\left(y^3\right)$$

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 $$\dfrac{{\displaystyle \lim_{u\to 8}\dfrac{u^{1/3}-8^{1/3}} {u-8}}} {{\displaystyle \lim_{v\to 1}\dfrac{v^{1/4}-1}{v-1}}}=\dfrac{\dfrac{1}{3}\cdot 8^{-2/3}}{\dfrac{1}{4}\cdot 1^{-3/4}} =\frac{1}{3}$$

By definition of derivative, by $$f(x)=\sqrt[3]{x+5}$$ and $$g(x)=\sqrt[4]{x-2}$$ we have

$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}=\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{x-3}\frac{x-3}{\sqrt[4]{x-2}-1}=\frac{f'(3)}{g'(3)}$$