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

If the value of the limit = \(e^{\lambda n}\). Find \(\lambda\)

My work:

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

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

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

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

How to calculate this limit?

## Comments

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TopNewestYou can use the fact that \( \displaystyle \lim_{x \rightarrow 0} f(x)^{g(x)} = \displaystyle \lim_{x \rightarrow 0} e^{g(x)(f(x) - 1)} \) – Siddhartha Srivastava · 2 years ago

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– Gautam Sharma · 2 years ago

ThanksLog in to reply

Answer Should be \(\lambda =3\) . It is \({ 1 }^{ \infty }\) form. – Deepanshu Gupta · 2 years ago

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– Gautam Sharma · 2 years ago

Yeah but explain.Log in to reply

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 \({ 1 }^{ \infty }\) form :

\(L=\lim _{ x\rightarrow a }{ { f\left( x \right) }^{ g\left( x \right) } } ={ e }^{ \lim _{ x\rightarrow a }{ { (f\left( x \right) -1)g\left( x \right) } } }\\ here:\quad \lim _{ x\rightarrow a }{ { f\left( x \right) } } =1\quad \& \quad \lim _{ x\rightarrow a }{ { g\left( x \right) } } =\infty \quad \) – Deepanshu Gupta · 2 years ago

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– Gautam Sharma · 2 years ago

Oh thanksLog in to reply