Limit problem involving cube roots (without use of L'Hôpital's rule or Taylor series)
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Let's change variable first: $u=x+8$.
Your limit becomes: $$ \lim_{u\rightarrow 0} \frac{(1+u)^{1/3}1+u}{1(12u)^{1/3}}. $$
Now use my comment above: $$ (1+u)^{1/3}1=\frac{u}{(1+u)^{2/3}+(1+u)^{1/3}+1} $$ and $$ 1(12u)^{1/3}=\frac{2u}{1+(12u)^{1/3}+(12u)^{2/3}} $$ Now your function becomes: $$ \frac{u(1+(12u)^{1/3}+(12u)^{2/3})}{2u((1+u)^{2/3}+(1+u)^{1/3}+1)} + \frac{u(1+(12u)^{1/3}+(12u)^{2/3})}{2u}. $$ 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, 2022Comments

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 yearsMaybe 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 yearsI meant to x tends to 8

Julien almost 10 years@17SI.34SA Yes it works.


André Nicolas almost 10 yearsI would recommend first letting $x=t8$. 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.