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Confusing Answers
- Thread starter Jason76
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Deleted member 4993
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- Indeterminate
OR
= 1
Both problems involve infinity to the 0 power. But they come out differently. Why?
Use the computer to graph those functions and investigate their behavior as those approach limit points.
Those are indeterminate forms. In the limit those approach different values
like
\(\displaystyle \displaystyle \lim_{x \to 2}\left [\dfrac{sin(x-2)}{x-2}\right]\) and \(\displaystyle \displaystyle \lim_{x \to 2}\left [\dfrac{x^2 - 4}{x-2}\right]\) both are of indeterminate form \(\displaystyle \frac{0}{0}\) but reach different values in the limit.
Both are indeterminate forms.
\(\displaystyle \liminf {x^{1/x}} = \infty ^{0}\) is indeterminate.
Do the same thing with this one you do with that other problem. It's a tool in your limit bag. Make its log a power of e. (Natural log and e undo each other. So if you do both, the value remains the same.)
\(\displaystyle \underset{x\rightarrow \infty}{ \lim } {x^{1/x}} = e^{\underset{x\rightarrow \infty}{ \lim } {{log{x^{1/x}}}}} = e^{\underset{x\rightarrow \infty}{ \lim } \frac{log(x)}{x}} = e^\frac{log(\infty)}{\infty} = e^{\frac{\infty}{\infty}}\) which is indeterminate.
Use L'Hospital for that limit. Take the derivative of the top and bottom. Derivative of logx is 1/x and derivative of x = 1.
\(\displaystyle e^{{\underset{x\rightarrow \infty}{ \lim } \frac{1/x}{1}}} = e^{{\underset{x\rightarrow \infty}{ \lim } \frac{1}{x}}}\)
The derivative of 1/x as x approaches infinity is 0.
\(\displaystyle e^{0} = 1\)
Step 1 is solve as you did. If it's indeterminate, then think about how to manipulate it, so you can use l'Hospital.
\(\displaystyle \liminf {x^{1/x}} = \infty ^{0}\) is indeterminate.
Do the same thing with this one you do with that other problem. It's a tool in your limit bag. Make its log a power of e. (Natural log and e undo each other. So if you do both, the value remains the same.)
\(\displaystyle \underset{x\rightarrow \infty}{ \lim } {x^{1/x}} = e^{\underset{x\rightarrow \infty}{ \lim } {{log{x^{1/x}}}}} = e^{\underset{x\rightarrow \infty}{ \lim } \frac{log(x)}{x}} = e^\frac{log(\infty)}{\infty} = e^{\frac{\infty}{\infty}}\) which is indeterminate.
Use L'Hospital for that limit. Take the derivative of the top and bottom. Derivative of logx is 1/x and derivative of x = 1.
\(\displaystyle e^{{\underset{x\rightarrow \infty}{ \lim } \frac{1/x}{1}}} = e^{{\underset{x\rightarrow \infty}{ \lim } \frac{1}{x}}}\)
The derivative of 1/x as x approaches infinity is 0.
\(\displaystyle e^{0} = 1\)
Step 1 is solve as you did. If it's indeterminate, then think about how to manipulate it, so you can use l'Hospital.