How is it limiting?

Calculus Level 3

A = lim ⁡ n → ∞ ( 1 + 1 n 2 + n ) n 2 + n A = \lim_{n\to \infty} \left ( 1 + \frac 1{n^2+n} \right)^{n^2 + \sqrt n }

What is the value of ln ⁡ ( A ) − 1 \ln(A)-1 ?


The answer is 0.

View wiki
Karan Arora by Karan Arora , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Raj Rajput
May 11, 2015

take log on both sides , R.H.S = (n square + root n ) / ( n square + n ) , and L.H.S = ln(A) , now solving R.H.S taking n square common from both numerator and denominator as limit is of infinite, making other terms in 1/n type form so when limit is applied to them they simply sort out to zero and value of ln(A) comes out to be 1 hence ln(A) -1 = 0

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Shilong Zhu
May 13, 2015

Call the expression inside the limit S, so then S=e^ln(S). Next use l'hopital's rule to solve the limit problem which will resolve to the power being 1, so e^1=S once limits are resolved. Hence, it is easy to see that ln(e)-1=0.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...