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

Limit:

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

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  • $\begingroup$ 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. $\endgroup$ – Julien Jan 29 '13 at 18:36
  • $\begingroup$ @julien: Not working! :( $\endgroup$ – 17SI.34SA Jan 29 '13 at 18:39
  • $\begingroup$ @17SI.34SA: Did you mean infinity or the 8 that you had in the limit? Regards $\endgroup$ – Amzoti Jan 29 '13 at 18:45
  • $\begingroup$ I meant to x tends to -8 $\endgroup$ – 17SI.34SA Jan 29 '13 at 18:55
  • $\begingroup$ @17SI.34SA Yes it works. $\endgroup$ – Julien Jan 29 '13 at 19:02
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Let's change variable first: $u=x+8$.

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

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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Hint: As $$\sqrt[3]a+\sqrt[3]b=\frac{a-b}{\sqrt[3]a^2-\sqrt[3]ab+\sqrt[3]b^2}$$ we have $$\lim_{x\to -8}\frac{\sqrt[3]{9+x}+x+7}{\sqrt[3]{15+2x}+1}=\lim_{x\to -8}\frac{9+x+(x+7)^3}{\sqrt[3]{9+x}^2-\sqrt[3]{9+x}(x+7)+(x+7)^2}\frac{\sqrt[3]{15+2x}^2-\sqrt[3]{15+2x}+1}{15+2x+1^3} = \lim_{x\to -8}\frac{9+x+(x+7)^3}{3}\frac{3}{16+2x}=\lim_{x\to -8}\frac{9+x+(x+7)^3}{16+2x} $$ Things should be straighforward from here on.

EDIT: The question was changed to $x\to -\infty$. Again everything reduces to computing $$\lim_{x\to -\infty}\frac{9+x+(x+7)^3}{16+2x}=\lim_{x\to -\infty}x^2\frac{\frac9{x^3}+\frac1{x^2}+(1+\frac7{x})^3}{\frac{16}x+2} $$ which is also simple.

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    $\begingroup$ 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. $\endgroup$ – André Nicolas Jan 29 '13 at 18:52

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