Limit of ln^2(x-2) as x approaches 3: Solving the Result

[sochdi]

hi, i want to ask help to solved my task.
if limit approach 3 of {ln^2(x-2)}^(x-3), how is the result?
thanks before

 

[Bohrok]

[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]

 

[sochdi]

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)...

 

[Bohrok]

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.

 

[sochdi]

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.

thanks for your answer.
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...

 

[Bohrok]

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.