limits of sequences

[calcnoob]

i have this problem here and the textbook shows how to solve it but im still lost as to why
here is the problem

and so far what i have

and don't know how to continue but the limit of the answer is e^something

 

[Deleted member 4993]

i have this problem here and the textbook shows how to solve it but im still lost as to why
here is the problem

and so far what i have

and don't know how to continue but the limit of the answer is e^something

Do you know L'Hopital's Rule?

 

[pka]

i have this problem here and the textbook shows how to solve it but im still lost as to why
here is the problem

and so far what i have

and don't know how to continue but the limit of the answer is e^something

If we define a sequence \(\displaystyle {e_n} = {\left( {1 + \frac{1}{n}} \right)^n}\) it is easy to show that sequence converges to \(\displaystyle e\)

Now in your sequence \(\displaystyle {\left( {1 - \frac{2}{n}} \right)^n}\) let \(\displaystyle \dfrac{1}{m}=\dfrac{-2}{n}\).

Then we can change your sequence to \(\displaystyle {\left[ {{{\left( {1 + \frac{1}{m}} \right)}^m}} \right]^{ - 2}}\)

Now it is easy to see that the limit is \(\displaystyle e^{-2}\).

 

[lookagain]

i have this problem here and the textbook shows how to solve it but im still lost as to why
here is the problem

and so far what i have

and don't know how to continue but the limit of the answer is e^something

calcnoob,

in the image, your second and third lines are incorrect.

In particular, the second line doesn't follow from the first line.

I think here is what is intended by you:


\(\displaystyle Present \ \ \ a_n \ = \ \bigg(1 - \dfrac{2}{n}\bigg)^n\)


The actual problem is to determine \(\displaystyle \ \displaystyle\lim_{n \to \infty} a_n.\)


Let y = \(\displaystyle \ \displaystyle\lim_{n \to \infty} a_n.\)


Or y = \(\displaystyle \ \displaystyle\lim_{n\to\infty} \bigg(1 - \dfrac{2}{n}\bigg)^n\)


ln(y) = \(\displaystyle \ ln\bigg[\ \displaystyle\lim_{n\to\infty} \bigg(1 - \dfrac{2}{n}\bigg)^n\bigg]\)


ln(y) = \(\displaystyle \ \displaystyle\lim_{n\to\infty} \ ln \bigg[\bigg(1 - \dfrac{2}{n}\bigg)^n \ \bigg]\)


ln(y) = \(\displaystyle \ \displaystyle\lim_{n\to\infty} \ \bigg\{ (n)ln \bigg(1 - \dfrac{2}{n}\bigg) \bigg\}\)


ln(y) = \(\displaystyle \ \displaystyle\lim_{n\to\infty} \ \dfrac{ln \bigg(1 - \dfrac{2}{n} \bigg)}{\bigg(\dfrac{1}{n}\bigg)}\)


Next, apply L'Hopital's Rule.


You'll arrive at:


ln(y) = something


Then have the constant e raised to each side:


\(\displaystyle e^{ln(y)} \ = \ e^{something}\)


Last, your equation will become in the form of:


\(\displaystyle y \ = \ e^{something}.\)


- - - - - - - - - - - - - - - - -


Edit:

If we define a sequence \(\displaystyle {e_n} = {\left( {1 + \frac{1}{n}} \right)^n}\) it is > > > easy ??? < < < to show that sequence converges to \(\displaystyle e\) Yes, for you.

That is a type of hand-waiving. The crux of that method would a challenge for the student to show the steps
that that sequence converges to e. Then the follow-up work/steps using the substitution to arrive at \(\displaystyle e^{-2}\)
would be significantly less difficult, as well as much less to write.