Limit

is the answer ∞

lim_{x→ ∞} (1 + 1/x )^x = e..
so the question reduces to
lim_{x→ ∞} ex/2 = ∞
Sir am i right?

it is of form
∞-∞
sensing something to do with expansion of
(1+1x)^x

@ xyz

how can it be ∞-∞ ????

e is a constant and as x--->∞

e= e itself

most probably it can be ∞ +∞ form

@govind i guess u cant do that

lim_{x→∞}((ex/2)+x²{(1+ 1/x)^x-e)} ≠lim_{x→∞} ex/2 +lim_{x→∞}x²{(1+ 1/x)^x-e)}

bec here respective limits r not finite

and btw ∞*0 is also an indeterminate form and its not equal to 0 ..as u hav did !

\\ \text{doing series expansion at x=}\infty,\text{as it converges to some value at x = }\infty \\ \left( 1+\frac{1}{x}\right)^{x}=1+1+\frac{1-\frac{1}{x}}{2!}+\frac{\left( {1-\frac{1}{x}\right)\left( 1-\frac{2}{x}\right)}} {3!}+.......\infty \\ \text{now we can split them as coeficients of }\frac{1}{x^n} \\ \text{the constant term turns out to be e}\\ \text{the coefficient of}\ \frac{1}{x} \ \text{comes out to be }\frac{-e}{2}\\ \text{and for calculating the coeficient of } \frac{1}{x^2} \text{we can make use of} \sum{\frac{(\sum{n})^2-\sum{n^2}}{2(n+1)!}}
hence
\left( 1+\frac{1}{x}\right)^{x}=1+1+\frac{1-\frac{1}{x}}{2!}+\frac{\left( {1-\frac{1}{x}\right)\left( 1-\frac{2}{x}\right)}} {3!}+.......\infty \\ \left( 1+\frac{1}{x}\right)^{x}=e-\frac{e}{2x}+\frac{11e}{24x^2}+\frac{k}{x^3}+..........\infty\\ or\\ \left( \left( 1+\frac{1}{x}\right)^{x}-e\right)=-\frac{e}{2x}+\frac{11e}{24x^2}+\frac{k}{x^3}+..........\infty \\ \text{mutiply by }x^2 \text{both sides }\\ x^2\left( \left( 1+\frac{1}{x}\right)^{x}-e\right)=-\frac{ex}{2}+\frac{11e}{24}+\frac{k}{x}+..........\infty \\ \boxed {\frac{ex}{2}+x^2\left( \left( 1+\frac{1}{x}\right)^{x}-e\right)=\frac{11e}{24}+0}(\because \ as \ x\rightarrow \infty \frac{1}{x}\rightarrow 0 )

That's correct xYz [1]

@rajat
\left( 1+\frac{1}{x}\right)^{x}=1+1+\frac{1-\frac{1}{x}}{2!}+\frac{\left( {1-\frac{1}{x}\right)\left( 1-\frac{2}{x}\right)}} {3!}+.......\infty \\ \text{if u see the coefficient of}\ \ \frac{1}{x} \\ =-\left(\frac{1}{2!}+\frac{1+2}{3!}+\frac{1+2+3}{4!}+...\infty \right)\\ =-\left(\sum_1^{\infty}{\frac{n(n+1)}{2(n+1)!}} \right)=-\left(\sum_1^{\infty}{\frac{1)}{2(n-1)!}} \right)=\frac{-e}{2} \\ \text{now the coefficient of }\frac{1}{x^2} \\ =\frac{1.2}{3!}+\frac{1.2+2.3+3.1}{4!}+.........\infty \\ =\sum{\frac{\left(\left( \frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2n+1)}{6}\right)}{2(n+1)!}} \\ \text{it reduces to } \\ \frac{1}{24}\sum{\left(\frac{3n}{(n-2)!}+\frac{2}{(n-2)!} \right)} \\ =\frac{1}{24}\left( 3*3e +2e \right)=\frac{11e}{24}

Look for the binomial expansion for an arbitrary real index. It goes like
(1+x)^\alpha =1+\alpha x +\dfrac{\alpha(\alpha-1)}{2!}x^2 + \dfrac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\ldots
where α is any real number.
It is easily seen that this series will terminate only if alpha is a positive integer. In all other cases the series does not terminate.