# Limit problem involving cube roots (without use of L'Hôpital's rule or Taylor series)

1,371

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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### 17SI.34SA

Updated on August 01, 2022

• 17SI.34SA 5 months

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

Limit:

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

• Julien almost 10 years
Maybe you could try using $a^{1/3}+b=(a+b^3)/(a^{2/3}-a^{1/3}b+b^2)$ for the numerator, and for the denominator.
• 17SI.34SA almost 10 years
@julien: Not working! :(
• Amzoti almost 10 years
@17SI.34SA: Did you mean infinity or the 8 that you had in the limit? Regards
• 17SI.34SA almost 10 years
I meant to x tends to -8
• Julien almost 10 years
@17SI.34SA Yes it works.
• André Nicolas almost 10 years
I would recommend first letting $x=t-8$. Our ugly expression becomes less ugly, and we are taking the limit as $t$ approaches $0$. Then one can use basically the algebraic trick above, but things "look" much nicer. Of course anyone really wanting to solve the problem would use Taylor series.