Help needed!!

I want help in a question that i came across limx0(r=0n(1)r.nCr(k=0nrnrCk2kxk)(x2x)r)1x\displaystyle\lim_{x \rightarrow 0}(\sum _{ r=0 }^{ n }{ (-1)^{ r }.^{ n }C_{ r } } (\sum _{ k=0 }^{ n-r }{ ^{ n-r }{ C }_{ k }{ 2 }^{ k }{ x }^{ k } } )(x^{ 2 }-x)^r) ^{\frac{1}{x}}

If the value of the limit = eλne^{\lambda n}. Find λ\lambda

My work:

limx0(r=0n(1)r.nCr(1+2x)nr(x2x)r)1x\Rightarrow\displaystyle\lim_{x \rightarrow 0}(\sum _{ r=0 }^{ n }{ (-1)^{ r }.^{ n }C_{ r } } (1+2x)^{n-r}(x^{ 2 }-x)^r)^ {\frac{1}{x}}

limx0(r=0nnCr(1+2x)nr(1)r(x2x)r)1x\Rightarrow\displaystyle\lim_{x \rightarrow 0}(\sum _{ r=0 }^{ n }{^{ n }C_{ r } } (1+2x)^{n-r}(-1)^{r}(x^{ 2 }-x)^r)^{\frac{1}{x}}

limx0(r=0nnCr(1+2x)nr(xx2)r)1x\Rightarrow \displaystyle\lim_{x \rightarrow 0}(\sum _{ r=0 }^{ n }{^{ n }C_{ r } } (1+2x)^{n-r}(x-x^2)^r)^{\frac{1}{x}}

limx0(1+3xx2)nx\Rightarrow\displaystyle\lim_{x \rightarrow 0}(1+3x-x^2)^{\frac{n}{x}}

How to calculate this limit?

Note by Gautam Sharma
4 years, 7 months ago

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1 vote

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Answer Should be λ=3\lambda =3 . It is 1{ 1 }^{ \infty } form.

Deepanshu Gupta - 4 years, 7 months ago

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change limit as L=enlimx0ln(1+3xx2)xL={ e }^{ n\lim _{ x\rightarrow 0 }{ \cfrac { \ln { (1+3x-{ x }^{ 2 }) } }{ x } } } now you can use expansion as suggested by taylor series.

Else this is 0/0 form so you can use L-hospital rule , if you had studied it yet.

Else There is direct formula for 1{ 1 }^{ \infty } form :

L=limxaf(x)g(x)=elimxa(f(x)1)g(x)here:limxaf(x)=1&limxag(x)=L=\lim _{ x\rightarrow a }{ { f\left( x \right) }^{ g\left( x \right) } } ={ e }^{ \lim _{ x\rightarrow a }{ { (f\left( x \right) -1)g\left( x \right) } } }\\ here:\quad \lim _{ x\rightarrow a }{ { f\left( x \right) } } =1\quad \& \quad \lim _{ x\rightarrow a }{ { g\left( x \right) } } =\infty \quad

Deepanshu Gupta - 4 years, 7 months ago

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Oh thanks

Gautam Sharma - 4 years, 7 months ago

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Yeah but explain.

Gautam Sharma - 4 years, 7 months ago

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You can use the fact that limx0f(x)g(x)=limx0eg(x)(f(x)1) \displaystyle \lim_{x \rightarrow 0} f(x)^{g(x)} = \displaystyle \lim_{x \rightarrow 0} e^{g(x)(f(x) - 1)}

Siddhartha Srivastava - 4 years, 7 months ago

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Thanks

Gautam Sharma - 4 years, 7 months ago

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