1is the answer ∞
Find:
\lim_{x\to\infty} \left(\dfrac{ex}{2}+x^2\left\{\left(1+\dfrac{1}{x}\right)^x-e\right\}\right)
Note that {.} is NOT the fractional part.
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29limx→ ∞ (1 + 1/x )x = e..
so the question reduces to
limx→ ∞ ex/2 = ∞
Sir am i right?
1@ xyz
how can it be ∞-∞ ????
e is a constant and as x--->∞
e= e itself
most probably it can be ∞ +∞ form
1@govind i guess u cant do that
limx→∞((ex/2)+x2{(1+ 1/x)x-e)} ≠limx→∞ ex/2 +limx→∞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 !
1\\ \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 )
1@xyz
can u plzz tell me that froom where did you get that expansion formula as i didn't know it..!
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}
1sorry- but i was asking about d expansion of (1+1/x)^x...
how did you got it..
i hav never even heard of it..!
66Look 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.