[Math] Limit problem involving cube roots (without use of L’Hôpital’s rule or Taylor series)

limits

How to find this limit without the help of L'Hôpital's rule nor expansion to Taylor series?

Limit:

$$\lim_{x\to -8}\frac{ (9+ x)^{1/3}+x+7}{(15+2 x)^{1/3}+1} $$

Best Answer

Let's change variable first: $u=x+8$.

Your limit becomes: $$ \lim_{u\rightarrow 0} \frac{(1+u)^{1/3}-1+u}{1-(1-2u)^{1/3}}. $$

Now use my comment above: $$ (1+u)^{1/3}-1=\frac{u}{(1+u)^{2/3}+(1+u)^{1/3}+1} $$ and $$ 1-(1-2u)^{1/3}=\frac{2u}{1+(1-2u)^{1/3}+(1-2u)^{2/3}} $$ Now your function becomes: $$ \frac{u(1+(1-2u)^{1/3}+(1-2u)^{2/3})}{2u((1+u)^{2/3}+(1+u)^{1/3}+1)} + \frac{u(1+(1-2u)^{1/3}+(1-2u)^{2/3})}{2u}. $$ Simplify the $u$'s and find that the limit is $1/2+3/2=2$.

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