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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
if limit approach 3 of {ln^2(x-2)}^(x-3), how is the result?
thanks before
no i cannot.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?
thanks for your answer.Sorry about the last post, I had something else in mind when I wrote that.
(ln^{2}(x - 2))^{x-3} = (ln(x - 2))^{2}^{(x-3)}, so
[tex](\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}}[/tex]
and
[tex]\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}}[/tex]
Now what you want to do is use l'Hôpital's rule to find that last limit.