Limits.

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

i encountered this problem

ProveProve thatthat

limx(a11x+a21x+.......an1xn)nx=a1a2.....an\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}}

aiR+a_i\in\mathbb{R^{+}}

MySolution:My Solution:

We re-write ai1x\large{a_i^{\frac{1}{x}}}=ai1x11x.1x+1\large{\dfrac{a_i^{\frac{1}{x}}-1}{\frac{1}{x}}.\frac{1}{x}+1}

doing this for all aia_i, our expression changes to:

(i=1nai1x11xnx+1)nx\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    1x0x\to\infty\implies\frac{1}{x}\to 0

for each aia_i : lim1x0ai1x11x=logeai\large{\lim_{\frac{1}{x}\to 0}\dfrac{a_i^{\frac{1}{x}}-1}{\frac{1}{x}}}=log_{e}a_{i}.......(by standard formula)

lim1x0i=1nai1x11x=i=1nlogeai=loge(a1a2....an)\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:

limx(1+loge(a1a2....an)nx)nx=eloge(a1a2....an)=a1a2....an\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 mathematicallymathematically correct?correct?

Note by Aritra Jana
4 years, 9 months ago

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1 vote

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@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 - 4 years, 9 months ago

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Sorry, er..., but I'm a -_- in calculus. Though I'm sure the calculus freak @Parth Lohomi would help.

Satvik Golechha - 4 years, 9 months ago

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:D okay then

Aritra Jana - 4 years, 9 months ago

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please don't tag me everywhere!! ;(

Parth Lohomi - 4 years, 9 months ago

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

Satvik Golechha - 4 years, 9 months ago

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@Satvik Golechha :v okay this was an awesome reply :v

Aritra Jana - 4 years, 9 months ago

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@Parth Lohomi oh c'mon..you cannot possibly deny that you are great at calculus!!

Aritra Jana - 4 years, 9 months ago

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@Sandeep Bhardwaj could you help? :D

Aritra Jana - 4 years, 9 months ago

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Yeah off course ! Whatever you did in your solution is absolutely correct. There is no mistake in that. There may be many different solutions to any problem with different approaches. Your approach is also good and correct. The general problems of this type (1) \left( 1^\infty \right) are done easily using the method :

Let you have to find the L=Limxa(f(x))g(x),whereLimxaf(x)=1andLimxag(x)=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, L=eLimxa(f(x)1)×g(x)\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 - 4 years, 9 months ago

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thanks a ton!!! :D

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 - 4 years, 9 months ago

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@Aritra Jana Not really L'Hospital's Rule . Because L'Hospitals Rule is applicable only for the forms of 00\dfrac{0}{0} and \dfrac{\infty}{\infty} . These limits are solved by taking the loglog both sides.

For eg. let y=Limxa(f(x))g(x),whereLimxaf(x)=1andLimxag(x)=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)×logf(x)log(y)=g(x) \times log f(x)

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

And Limxag(x)×logf(x) Lim_{x \to a} g(x) \times logf(x) is of the form ×0 \infty \times 0 now, which can be further converted into 00\dfrac{0}{0} and \dfrac{\infty}{\infty} forms , and after that you can apply L'Hospitals Rule.

Suppose limxag(x)×logf(x)=λlim_{x \to a} g(x) \times logf(x) = \lambda

    y=Limxa(f(x))g(x)=eλ\implies y= \displaystyle Lim_{x \to a} (f(x))^{g(x)}=e^\lambda.

Sandeep Bhardwaj - 4 years, 9 months ago

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@Sandeep Bhardwaj ohh.. Thanks for the clarification! :D

Aritra Jana - 4 years, 9 months ago

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