# Limits.

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 $mathematically$ $correct?$ Note by Aritra Jana
6 years, 8 months ago

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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

- 6 years, 8 months ago

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

- 6 years, 8 months ago

:D okay then

- 6 years, 8 months ago

please don't tag me everywhere!! ;(

- 6 years, 8 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.

- 6 years, 8 months ago

:v okay this was an awesome reply :v

- 6 years, 8 months ago

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

- 6 years, 8 months ago

@Sandeep Bhardwaj could you help? :D

- 6 years, 8 months ago

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 $\left( 1^\infty \right)$ are done easily using the method :

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

- 6 years, 8 months ago

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?

- 6 years, 8 months ago

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

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

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

- 6 years, 8 months ago

ohh.. Thanks for the clarification! :D

- 6 years, 8 months ago