Limit Question with e^

[markraz]

Hi,

I'm trying calculate a limit (doing an Infinite series root test )

\(\displaystyle \Large\lim_{n \to \infty}( \frac{2}{n^\frac{5}{n}})\)

\(\displaystyle \Large\lim_{n \to \infty}( \frac{2}{n^\frac{5}{n}})\) => \(\displaystyle \Large\lim_{x \to \infty} \frac{2}{ e^{\frac{5}{x} ln(x)} }\)

(Work on just Denominator for clarity)

\(\displaystyle \Large\lim_{x \to \infty} e^{\frac{5}{x} ln(x)}\)

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\(\displaystyle \Large\lim_{x \to \infty} \Large e^{\frac{5 ln(x)}{x}} \) do I do L'Hôpital's rule here with the \(\displaystyle \Large {\frac{5 ln(x)}{x}} \)?


\(\displaystyle \Large{\frac{5 ln(x)}{x}} \) => L'Hôpital's = \(\displaystyle \Large {5 (\frac {\frac{1}{x}} {1})} \) = \(\displaystyle \Large5 {\frac{1} {\infty}} \)= (5 * 0) = 0 ????

\(\displaystyle \Large\lim_{x \to \infty} e^0\) = 1

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finally put original back solved Denominator with Numerator.

\(\displaystyle \Large\frac{2}{e^0}\) = \(\displaystyle \Large\frac{2}{1}\) = 2

is the L'Hôpital's approach valid?? or is there a more efficient /correct way to solve ?? \(\displaystyle \Large\lim_{n \to \infty}( \frac{2}{n^\frac{5}{n}})=2\)
what approach is optimal?

Thanks

 

[Ishuda]

Not more correct (although some of your steps would need to be justified depending on the context of the question). A very similar approach is to take logs, do the limit and then do the exponential back. It also would have some steps needing justifications depending on context.

 

[markraz]

Not more correct (although some of your steps would need to be justified depending on the context of the question). A very similar approach is to take logs, do the limit and then do the exponential back. It also would have some steps needing justifications depending on context.

Thanks, so the L'Hôpital approach is valid here? and I actually solved it correctly?
sometimes I do problems wrong, but get the right answer and my instructor marks them wrong

thanks

 

[pka]

\(\displaystyle \Large\lim_{n \to \infty}( \frac{2}{n^\frac{5}{n}})\)

All you need to note is: \(\displaystyle \Large\displaystyle{\lim _{n \to \infty }}\sqrt[n]{{{n^5}}} = 1\)

 

[Ishuda]

Thanks, so the L'Hôpital approach is valid here? and I actually solved it correctly?
sometimes I do problems wrong, but get the right answer and my instructor marks them wrong

thanks

Yes, you solved it correctly. However 'doing problems wrong' is part of what I was alluding to when I said possibly some justifications had to be made. What justifies taking limits like that. That is what justifies if a = lim a_n, then lim e^a_n = e^a. Depending on just where you are (what level) and what the test is about, your instructor may not need/want this step as it is 'a given' to wanting/needing a full blown proof of that step. Typically, I had found that it was 'a given'.