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I am trying to evaluate this limit as follows, $$\lim_{x\to 2^+} (x-2)^{x^2-4} = e^{\lim_{x\to 2^+} x^2-4 \cdot \ln(x-2)}$$

Aside, we can apply L'Hospital to the limit in the exponent by rearranging to form $\frac{-\infty}{\infty}$: $$\lim_{x\to 2^+} \frac{\ln(x-2)}{\frac{1}{x^2-4}} = \lim_{x\to 2^+}\frac{\frac{1}{x-2}}{\frac{2x}{(x^2-4)^2}}$$

At this point I'm kind of lost as to what I should do. I tried rearranging the fraction and then simplifying as follows and applying L'Hospital again since we have $\frac{0}{0}$ $$=\lim_{x\to 2^+}\frac{(x^2-4)^2}{(x-2)(2x)} = \lim_{x\to 2^+}\frac{2(x^2-4)\cdot2x}{4x-4} =\frac{0}{4} =0 $$ I am supposed to get $1$ as the answer. I have a feeling I might be doing some sort of illegal manipulation, because I've somehow managed to make this limit equal to $4$ and $8$ as well using similiar techniques. What am I doing wrong?

Am I allowed to restructure the limit as I did after applying L'Hospital's rule once? If I keep it in the same fraction over another fraction form and apply L'Hospital's rule again, I get something even messier.

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You did alright, you just forget the exponential outside. $\exp(0)=1$.

Alternatively, note that

$$\lim_{x \to 2^+}\frac{(x^2-4)^2}{(x-2)(2x)}=\lim_{x \to 2^+}\frac{(x-2)^2(x+2)^2}{(x-2)(2x)}=\lim_{x \to 2^+}\frac{(x-2)(x+2)^2}{(2x)}=0$$

then remember to take exponential.

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You may substitute $t = x-2$ and consider $t \rightarrow 0^+$ and use

  • $t^t \stackrel{t \to 0^+}{\longrightarrow} 1$ and note that
  • $x^2-4 = (x-2)(x+2)$

$$(x-2)^{x^2-4} = \left( t^t\right)^{t+4} \stackrel{t \to 0^+}{\longrightarrow} 1^4 = 1$$

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Let $ y=(x-2)^{x^2-4}$, then $ \log y=(x^2-4) \log (x-2)=\frac{\log(x-2)}{\frac{1}{x^2-4}}$.

Now $ \lim_{x \to 2^{+}} \log y= \lim_{x \to 2^{+}} \frac{\log(x-2)}{\frac{1}{x^2-4}}= \ \left( \frac{\infty}{\infty} \right) \ form $

By L'Hospital's rule

$ \lim_{x \to 2^{+}} \log y= \lim_{x \to 2^{+}} \frac{(x^2-4)^2}{-2x(x-2)}=\lim_{x \to 2^{+}} \frac{(x-2)(x+2)^2}{-2x}=0 $

$ \Rightarrow \lim_{x \to 2^{+}} \log y=0=\log 1=\lim_{x \to 2^{+}} \log 1 \\ \Rightarrow \lim_{x \to 2^{+}} y=1 \\ \Rightarrow \lim_{x \to 2^{+}}(x-2)^{x^2-4}=1 $

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We have that

$$(x-2)^{x^2-4}=e^{(x^2-4)\log(x-2)} \to 1$$

indeed by standard limits by $t=x-2 \to 0^+$

$$(x^2-4)\log(x-2)=(t+4)\cdot t\log t \to 4\cdot0=0$$

indeed by $y=\frac1t \to \infty$

$$t\log t=\frac1y \log \left(\frac1y\right)=-\frac{\log y}{y} \to 0$$

and the last we can refer to the related

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