Limit Problem - # 2
- Thread starter Jason76
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\(\displaystyle \displaystyle \lim_{x \to \infty}\left( x^{4/x}\right)\)\(\displaystyle \lim x \rightarrow \infty\)
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle \infty^{\dfrac{4}{\infty}}\)
\(\displaystyle \infty^{0}\)What does this become
Since \(\displaystyle x\) occurs in an exponent, consider the logarithm:
\(\displaystyle \displaystyle \lim_{x \to \infty} \left(\dfrac{4}{x}\ln x\right) = \lim_{x \to \infty} \dfrac{4\ \ln x}{x} \)
Can you find the limit of the logarithm? Then what is limit of the original expression?
It comes out to Indeterminate \(\displaystyle \dfrac{\infty}{\infty}\) so maybe use L Hopital's rule.
Why "maybe"?
\(\displaystyle \lim x \rightarrow \infty\)
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle \infty^{\dfrac{4}{\infty}}\)
\(\displaystyle \infty^{0}\)
Quotient Rule or not - I think not.
With Quotient Rule
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4\ln x}{x}\)
\(\displaystyle \dfrac{(x)(\dfrac{4}{x} - \ln x (1)}{x^{2}}\)
\(\displaystyle \dfrac{4 (\infty) - 4 \ln (\infty)}{\infty^{2}}\)
Without Quotient
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4 \ln x}{x}\)
\(\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}\)
\(\displaystyle \dfrac{4 \ln (\infty)}{(\infty)}\)
\(\displaystyle \dfrac{(\infty)}{(\infty)} = \) Indeterminate What now?
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle \infty^{\dfrac{4}{\infty}}\)
\(\displaystyle \infty^{0}\)
Quotient Rule or not - I think not.
With Quotient Rule
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4\ln x}{x}\)
\(\displaystyle \dfrac{(x)(\dfrac{4}{x} - \ln x (1)}{x^{2}}\)
\(\displaystyle \dfrac{4 (\infty) - 4 \ln (\infty)}{\infty^{2}}\)
Without Quotient
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4 \ln x}{x}\)
\(\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}\)
\(\displaystyle \dfrac{4 \ln (\infty)}{(\infty)}\)
\(\displaystyle \dfrac{(\infty)}{(\infty)} = \) Indeterminate
What now?
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You must ALWAYS differentiate first and evaluate last.\(\displaystyle \lim x \rightarrow \infty\)
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle \infty^{\dfrac{4}{\infty}}\)
\(\displaystyle \infty^{0}\)
Quotient Rule or not - I think not.
With Quotient Rule
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4\ln x}{x}\)
\(\displaystyle \dfrac{(x)(\dfrac{4}{x} - \ln x (1)}{x^{2}}\)
\(\displaystyle \dfrac{4 (\infty) - 4 \ln (\infty)}{\infty^{2}}\)
Without Quotient
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4 \ln x}{x}\)
\(\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}\)... DO the differentiation!
\(\displaystyle \dfrac{4 \ln (\infty)}{(\infty)}\)
\(\displaystyle \dfrac{(\infty)}{(\infty)} = \) Indeterminate
After you find the limit of the logarithm, what is the limit of the original expression?
In the end it evaluates to 0, because \(\displaystyle \dfrac{4}{x}\) comes on top, and evaluates to \(\displaystyle 0\), and the bottom become 1.
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Yes, "IT" evaluates to 0 in the limit -- but what is "IT"? You need to be a little more explicit.
And how does "IT" being zero relate to the original question?
If you need a hint, see my first response, post #2 in this thread.
\(\displaystyle \ln x \rightarrow \infty\)
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle (\infty)^{\dfrac{4}{\infty}}\)
\(\displaystyle (\infty)^{0}\) - Indeterminate
\(\displaystyle x^{\dfrac{4}{x}}\) - Rewrite
\(\displaystyle \dfrac{4}{x} \ln x\)
\(\displaystyle \dfrac{4 \ln x}{x} \)
\(\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x} \)
\(\displaystyle \dfrac{\dfrac{4}{x}}{1}\)
\(\displaystyle \dfrac{\dfrac{4}{\infty}}{1}\)
\(\displaystyle \dfrac{0}{1} = 0\) Answer
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle (\infty)^{\dfrac{4}{\infty}}\)
\(\displaystyle (\infty)^{0}\) - Indeterminate
\(\displaystyle x^{\dfrac{4}{x}}\) - Rewrite
\(\displaystyle \dfrac{4}{x} \ln x\)
\(\displaystyle \dfrac{4 \ln x}{x} \)
\(\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x} \)
\(\displaystyle \dfrac{\dfrac{4}{x}}{1}\)
\(\displaystyle \dfrac{\dfrac{4}{\infty}}{1}\)
\(\displaystyle \dfrac{0}{1} = 0\)
AnswerHallsofIvy
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Do you not remember what the question was?
You were asked to find the limit, as x goes to infinity, of \(\displaystyle y= x^{\dfrac{4}{x}}\).
Since you cannot do that directly, you take the logarithm: \(\displaystyle ln(y)= ln(x^{\frac{4}{x}}= \dfrac{4ln(x)}{x}\).
Those both go to "infinity" as x goes to "infinity" so you use "L'Hopital's rule" differentiating both numerator and denomintor. The derivtive of 4 ln(x) is 4/x and the derivative of x is 1. So now we re looking at
\(\displaystyle \lim_{x\to\infty} ln(y)= \lim_{x\to\infty}\dfrac{4 ln(x)}{x}\)\(\displaystyle = \lim_{x\to\infty}\dfrac{\dfrac{4}{x}}{1}= \lim_{x\to\infty} \dfrac{4}{x}= 0\).
So you have shown that \(\displaystyle \lim_{x\to\infty} ln(y)= 0\). Now, what does that tell you about your original problem? Do you see what Dr. Phil meant when he asked "what does IT mean?"
You were asked to find the limit, as x goes to infinity, of \(\displaystyle y= x^{\dfrac{4}{x}}\).
Since you cannot do that directly, you take the logarithm: \(\displaystyle ln(y)= ln(x^{\frac{4}{x}}= \dfrac{4ln(x)}{x}\).
Those both go to "infinity" as x goes to "infinity" so you use "L'Hopital's rule" differentiating both numerator and denomintor. The derivtive of 4 ln(x) is 4/x and the derivative of x is 1. So now we re looking at
\(\displaystyle \lim_{x\to\infty} ln(y)= \lim_{x\to\infty}\dfrac{4 ln(x)}{x}\)\(\displaystyle = \lim_{x\to\infty}\dfrac{\dfrac{4}{x}}{1}= \lim_{x\to\infty} \dfrac{4}{x}= 0\).
So you have shown that \(\displaystyle \lim_{x\to\infty} ln(y)= 0\). Now, what does that tell you about your original problem? Do you see what Dr. Phil meant when he asked "what does IT mean?"
\(\displaystyle \lim x \rightarrow \infty\)
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle \infty^{\dfrac{4}{\infty}}\)
\(\displaystyle \infty^{0}\) Indeterminate
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4 \ln x}{x}\)
\(\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}\)
\(\displaystyle \dfrac{\dfrac{4}{x}}{1}\)
\(\displaystyle \dfrac{\dfrac{4}{\infty}}{1} = 0\) Answer - computer says this is wrong. Did I overlook something?
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle \infty^{\dfrac{4}{\infty}}\)
\(\displaystyle \infty^{0}\) Indeterminate
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4 \ln x}{x}\)
\(\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}\)
\(\displaystyle \dfrac{\dfrac{4}{x}}{1}\)
\(\displaystyle \dfrac{\dfrac{4}{\infty}}{1} = 0\)
Answer - computer says this is wrong. Did I overlook something?
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That is NOT the answer. What was the question, and what have you found? Do you EVER read what we tell you? Look back at recent posts by HallsofIvy and by me.\(\displaystyle \lim x \rightarrow \infty\)
\(\displaystyle x^{\dfrac{4}{x}}\)
\(\displaystyle \infty^{\dfrac{4}{\infty}}\)
\(\displaystyle \infty^{0}\) Indeterminate - and yet, ANY number raised to the power 0 is 1
\(\displaystyle \ln[x^{\dfrac{4}{x}}]\)
\(\displaystyle \dfrac{4 \ln x}{x}\)
\(\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}\)
\(\displaystyle \dfrac{\dfrac{4}{x}}{1}\)
\(\displaystyle \dfrac{\dfrac{4}{\infty}}{1} = 0\)
I was a bit surprised to notice that - HallsofIvy will tell us whether \(\displaystyle \infty^0 = 1\) is rigorous enough to be acceptable as the answer.1 is the answer without having to use L' Hopital's
What we have been (vainly) TRYING to get you to do is see that 0 is the limit for the logarithm, and \(\displaystyle e^{\ln x} = e^0 = 1\) is the final answer for the limit of the function.
DrPhil, let me ask. Are you asking about that form in general?
Look at these examples where each would be substituted into \(\displaystyle \ \displaystyle\lim_{x \to \infty}a^b \ \ to \ \ get \ \ the \ \ \infty^0 \ \ respective \ \ indeterminate \ \ form:\)
\(\displaystyle a = e^{x^2}, \ \ \ b = \dfrac{1}{x}, \ \ \ limit = \infty\)
\(\displaystyle a = e^x, \ \ \ b = \dfrac{1}{x}, \ \ \ limit = e\)
\(\displaystyle a = e^x, \ \ \ b = \dfrac{1}{x^2}, \ \ \ limit = 1\)
OK - that wraps it up .. depends on how the infinity was generated.
Jason - no shortcut! l'Hospital is required.
Thanks lookagain!