I want to know if the following thing can be done in a limit.

i encountered this problem

\(Prove\) \(that\)

\[\large{\lim_{x\to\infty}(\dfrac{a_1^{\frac{1}{x}}+a_2^{\frac{1}{x}}+.......a_n^{\frac{1}{x}}}{n})^{nx}=a_{1}a_{2}.....a_{n}}\]

\(a_i\in\mathbb{R^{+}}\)

\(My Solution:\)

We re-write \(\large{a_i^{\frac{1}{x}}}\)=\(\large{\dfrac{a_i^{\frac{1}{x}}-1}{\frac{1}{x}}.\frac{1}{x}+1}\)

doing this for all \(a_i\), our expression changes to:

\[\large{(\dfrac{\sum\limits_{i=1}^{n}{\dfrac{a_i^{\frac{1}{x}}-1}{\frac{1}{x}}}}{nx}+1)^{nx}}\]

Now here's the part i don't think is correct:

since \(x\to\infty\implies\frac{1}{x}\to 0\)

for each \(a_i\) : \(\large{\lim_{\frac{1}{x}\to 0}\dfrac{a_i^{\frac{1}{x}}-1}{\frac{1}{x}}}=log_{e}a_{i}\).......(by standard formula)

\(\therefore \large\lim_{\frac{1}{x}\to 0} \large{\sum\limits_{i=1}^{n}{\dfrac{a_i^{\frac{1}{x}}-1}{\frac{1}{x}}}=\sum\limits_{i=1}^{n}{log_{e}a_i}=log_{e}(a_{1}a_{2}....a_{n})}\)

so, we have our expression as:

\[\lim_{x\to\infty}\large{(1+\dfrac{log_{e}(a_{1}a_{2}....a_{n})}{nx})^{nx}}=\large{e^{log_{e}(a_{1}a_{2}....a_{n})}}=a_{1}a_{2}....a_{n}\]

as required.

are the steps correct? what I mean to ask is that: can we derive the limits of the different parts independently and put their values to derive the limit of that function. 1 is this \(mathematically\) \(correct?\)

## Comments

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TopNewest@Sandeep Bhardwaj could you help? :D – Aritra Jana · 2 years, 4 months ago

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Let you have to find the \(L= \displaystyle Lim_{x \to a} (f(x))^{g(x)}, \quad where \quad Lim_{x \rightarrow a} f(x)=1 \quad and \quad Lim_{x \rightarrow a} g(x) =\infty\).

then, \[\displaystyle L= e^{Lim_{x \rightarrow a}(f(x)-1) \times g(x)}\]

You are doing great. I appreciate your this innovative approach to learn something. Very good. Keep it up. !!!

PS : The difference between a master and a beginner is that the master has failed as many times as the beginner has not even tried. !@Aritra Jana – Sandeep Bhardwaj · 2 years, 4 months agoLog in to reply

You know this means a lot!!! :D

and thanks for that extra bit of info on these types of limits! is that a part of L'Hospitals rule? – Aritra Jana · 2 years, 4 months ago

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For eg. let \(y= \displaystyle Lim_{x \to a} (f(x))^{g(x)}, \quad where \quad Lim_{x \rightarrow a} f(x)=1 \quad and \quad Lim_{x \rightarrow a} g(x) =\infty\).

taking log both sides\(log(y)=g(x) \times log f(x)\)

Now we solve the limit for \(log(y)\)..!!

And \( Lim_{x \to a} g(x) \times logf(x)\) is of the form \( \infty \times 0\) now, which can be further converted into \(\dfrac{0}{0}\) and \(\dfrac{\infty}{\infty}\) forms , and after that you can apply L'Hospitals Rule.

Suppose \(lim_{x \to a} g(x) \times logf(x) = \lambda\)

\(\implies y= \displaystyle Lim_{x \to a} (f(x))^{g(x)}=e^\lambda\). – Sandeep Bhardwaj · 2 years, 4 months ago

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– Aritra Jana · 2 years, 4 months ago

ohh.. Thanks for the clarification! :DLog in to reply

@Parth Lohomi @Calvin Lin @Mursalin Habib @megh choksi @Satvik Golechha @HARSH SHrivastava i have just started learning about limits. so if there are any gross mistakes i apologize beforehand.

it would mean a great deal if you helped me out :D – Aritra Jana · 2 years, 4 months ago

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-_-in calculus. Though I'm sure the calculus freak @Parth Lohomi would help. – Satvik Golechha · 2 years, 4 months agoLog in to reply

– Aritra Jana · 2 years, 4 months ago

:D okay thenLog in to reply

– Parth Lohomi · 2 years, 4 months ago

please don't tag me everywhere!! ;(Log in to reply

– Satvik Golechha · 2 years, 4 months ago

Sorry, thought that you'd like to aid others, howbeit thy prodigiously phenomenal and savvy calculus endowment presumably impelled me to tag you.Log in to reply

– Aritra Jana · 2 years, 4 months ago

:v okay this was an awesome reply :vLog in to reply

– Aritra Jana · 2 years, 4 months ago

oh c'mon..you cannot possibly deny that you are great at calculus!!Log in to reply