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I need some help in calculating this limit:

$\lim_{x\rightarrow2}(x-1)^{\frac{2x^2-8}{x^2-4x+4}}$

Thanks a lot.

Jan
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Adi Cohen
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  • at x=2 !! sorry for not writing in main passage – Adi Cohen Feb 14 '17 at 17:35
  • Is the limit of interest given by $\lim_{x\to 2}(x-1)^{(2x^2-8)/(x^2-4x+4)}$? – Mark Viola Feb 14 '17 at 17:36
  • What have you tried? Where are you stuck? Hint: calculate the limit of the power first, using L'Hopital's law. Does not work since the limit is $\pm\infty$? Then transform your expression via $\log(a^b)=b\log(a)$. Then if $b\rightarrow\infty$, $1/b\rightarrow0$. – Jan Feb 14 '17 at 17:40

2 Answers2

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It's probably better to do a change of variable, $x=t+2$, so the exponent becomes $$ \frac{2x^2-8}{x^2-4x+4}=\frac{2(x-2)(x+2)}{(x-2)^2}=2\frac{x+2}{x-2}=2\frac{t+4}{t} $$ Thus the limit becomes $$ \lim_{t\to0}\bigl((1+t)^{1/t}\bigr)^{2(t+4)} $$ Now it's a matter of computing $$ l=\lim_{t\to0}(1+t)^{1/t} $$ and then the sought limit is $l^8$.

egreg
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HINTS:

First, note that

$$\frac{2x^2-8}{x^2-4x+4}=\frac{2(x+2)}{x-2}$$

Hence, we can write

$$(x-1)^{\frac{2(x+2)}{x-2}}=e^{\frac{2(x+2)}{x-2}\log(x-1)}$$

Finish by using L'Hospital's Rule to show

$$\lim_{x\to 2}\frac{2(x+2)}{x-2}\log(x-1)=\lim_{x\to 2}\left(2\log(x-1)+\frac{2(x+2)}{x-1}\right)$$

Mark Viola
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