some limits

[nutzu77]

Hello! Can someone explain me how to solve those limits? (i want just hints, not full solved )

1. lim (1-1/4)*(1-1/9)*...*(1-1/n^2) with n->inf
2. lim(√(n^2+2n+5)-n) , n-> inf
3. lim n*ln(√(n^2+2n+5)-n) , n-> inf

Thanks !

 

[Deleted member 4993]

Hello! Can someone explain me how to solve those limits? (i want just hints, not full solved )

1. lim (1-1/4)*(1-1/9)*...*(1-1/n^2) with n->inf
2. lim(√(n^2+2n+5)-n) , n-> inf
3. lim n*ln(√(n^2+2n+5)-n) , n-> inf

Thanks !

1. (1-1/4)*(1-1/9)*...*(1-1/n^2)

= (1 + 1/2)(1 - 1/2)(1+1/3)(1 -1/3)(1+1/4)(1-1/4) ...... (1+1/n)(1-1/n)

= (1 + 1/2)(1+1/3)(1+1/4) ...... (1+1/n) * (1 - 1/2)(1 -1/3)(1-1/4) ...... (1-1/n)

= (3/2)(4/3)(5/4) ....[(n+1)/n] * (1/2)(2/3)(3/4) ..... [(n-1)/n]

= (n+1)/2 *1/n ..... Continue..[edited]

You start the rest....

 

[Ishuda]

Hello! Can someone explain me how to solve those limits? (i want just hints, not full solved )

1. lim (1-1/4)*(1-1/9)*...*(1-1/n^2) with n->inf
2. lim(√(n^2+2n+5)-n) , n-> inf
3. lim n*ln(√(n^2+2n+5)-n) , n-> inf

Thanks !

What can you use? For example, suppose b=lim[ln(a_n)], could you use lim[e^a_n] = e^b?

How detailed do you need to get? What kind of justification would you have to give to expand the square root say
\(\displaystyle \sqrt{n^2+1} = n \sqrt{1+\frac{1}{n^2}} \) ~ \(\displaystyle n [1 + \frac{1}{2n^2} - \frac{1}{8 n^4} + ... \) ~ \(\displaystyle n + \frac{1}{2n} - \frac{1}{8 n^3} + ...\)
and take limits across

Or to expand the natural log function.

 

[stapel]

1. lim (1-1/4)*(1-1/9)*...*(1-1/n^2) with n->inf

I think the above is as follows:

. . . . .\(\displaystyle \displaystyle{ \lim_{n\, \rightarrow\, \infty} \, \left[ \left(1\, -\, \frac{1}{4}\right) \cdot \left(1\, -\, \frac{1}{9}\right) \cdot\, ...\, \cdot \left(1\, -\, \frac{1}{n^2}\right) \right] }\)

Convert the subtractions into fractions:

. . . . .\(\displaystyle \displaystyle{ 1\, -\, \frac{1}{k^2}\, =\, \frac{k^2\, -\, 1}{k^2}\, =\, \frac{(k\, -\, 1)(k\, +\, 1)}{k^2} }\)

Then note that, at each stage (other than at the "ends"), the k - 1 will cancel with the denominator to the left and the k + 1 will cancel with the denominator to the right. What will be left?

2. lim(√(n^2+2n+5)-n) , n-> inf

Use the customary "trick" of multiplying, top and bottom, by the conjugate:

. . . . .\(\displaystyle \displaystyle{ \left(\frac{\sqrt{n^2\, +\, 2n\, +\, 5\, } \, -\, n}{1}\right)\left(\frac{\sqrt{n^2\, +\, 2n\, +\, 5\, }\, +\, n}{\sqrt{n^2\, +\, 2n\, +\, 5\, }\, +\, n}\right) }\)

Then simplify and take the limit.