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sochdi
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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
Bohrok said: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?
Bohrok said: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
[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.
"Limit ln^2" refers to the mathematical concept of taking the limit of the natural logarithm squared of a function as the input of the function approaches a certain value.
"Limit ln^2" is important in mathematics because it allows us to analyze the behavior of functions as they approach a specific value. This can help us understand the properties and characteristics of the function and make predictions about its behavior.
The limit of ln^2 can be calculated by first taking the natural logarithm of the function, then squaring the result, and finally taking the limit as the input of the function approaches the desired value.
The limit of ln^2 and the derivative are closely related. In fact, the derivative of a function can be expressed as the limit of ln^2 as the input of the function approaches 0. This relationship is known as the definition of the derivative.
Yes, "Limit ln^2" has many real-life applications in fields such as physics, economics, and engineering. For example, it can be used to model the growth rate of a population or the decay of a radioactive substance.