# Limit ln^2

if limit approach 3 of {ln^2(x-2)}^(x-3), how is the result?
thanks before

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[STRIKE]As it's written, {ln^2(x-2)}^(x-3) is continuous for x>2, so you can just plug in x=3 to get the limit. Is that exactly what the original expression was?[/STRIKE]

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As it's written, {ln^2(x-2)}^(x-3) is continuous for x>2, so you can just plug in x=3 to get the limit. Is that exactly what the original expression was?
no i cannot.
if i plug in x=3, so it will be {ln^2(3-2)}^(3-3) = {ln^2 1}^0 = 0^0 (unconditional)...

Sorry about the last post, I had something else in mind when I wrote that.

(ln2(x - 2))x-3 = (ln(x - 2))2(x-3), so

$$(\ln^2(x - 2))^{x-3} = e^{\ln(\ln(x-2))^{2(x-3)}} = e^{2(x-3)\ln(\ln(x-2))} = e^{2\cdot\frac{\ln(\ln(x-2))}{\frac{1}{x-3}}$$
and
$$\lim_{x\rightarrow 3}(\ln^2(x - 2))^{x-3} = e^{2\lim_{x\rightarrow 3}\frac{\ln(\ln(x-2))}{\frac{1}{x-3}}$$

Now what you want to do is use l'Hôpital's rule to find that last limit.

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Sorry about the last post, I had something else in mind when I wrote that.

(ln2(x - 2))x-3 = (ln(x - 2))2(x-3), so

$$(\ln^2(x - 2))^{x-3} = e^{\ln(\ln(x-2))^{2(x-3)}} = e^{2(x-3)\ln(\ln(x-2))} = e^{2\cdot\frac{\ln(\ln(x-2))}{\frac{1}{x-3}}$$
and
$$\lim_{x\rightarrow 3}(\ln^2(x - 2))^{x-3} = e^{2\lim_{x\rightarrow 3}\frac{\ln(\ln(x-2))}{\frac{1}{x-3}}$$

Now what you want to do is use l'Hôpital's rule to find that last limit.
anyway, may i know why (ln2(x - 2))(x-3) can become (ln(x - 2))2(x-3)?

because (ln2(x - 2)) = {ln(x-2) . ln(x-2)} and it's not same with {ln(x-2)}2 right?

@others any other idea please show me...

ln2x = (ln x)2, so as an example,

(ln2x)a = ((ln x)2)a = (ln x)2·a = (ln x)2a

just using properties of exponents.

HallsofIvy
For any function f, $f^2(x)$ is defined as $(f(x))^2$ so, yes, they are the same thing.