1lnL=\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{r=1}^{n}{\frac{r}{n}ln\frac{r}{n}}
lnL=\int_{0}^{1}{x\: lnx\: dx}=-\frac{1}{4}
L=e^{-\frac{1}{4}}
7 Answers
1
49sir can u elaborate a bit more??plzzz...[1]
no latex is working again....refresh page avik 4 better use...[1]
13Thanks Bipin Bhaiya [1]
@atgs... just take log both sides & expand, it will reduce to the form of limit of a sum, fir simple integral hai. [1]
132 more, might be simple, but am not getting them...
2) Lim(n→∞) (n+4)!/n!
3) Lim(x→0) x2.Sin(1/x)
P.S. I do not have the answers.
1is the ans -- 0
my [ mostly wrong ] metod
\lim_{x\rightarrow 0} x^2 .sin\left(\frac{1}{x} \right)
\lim_{x\rightarrow 0} x^2 .\frac{sin\left(\frac{1}{x} \right)\frac{1}{x}}{\frac{1}{x}}
\lim_{x\rightarrow 0} x = 0
1ya thats wrong
limx→a sinf(x)f(x) is only eqaul to 1 wen f(x)→0 wen x→a
here 1/x →∞ wen x→0
13)
lim = 0
x2→0 while Sin(1/x) is bounded between [-1,1].
thus limit is zero .
2) lim (n→∞) (n+4)!n! = ∞