# Help needed!!

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? Note by Gautam Sharma
5 years, 8 months ago

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

- 5 years, 8 months ago

change limit as $L={ e }^{ n\lim _{ x\rightarrow 0 }{ \cfrac { \ln { (1+3x-{ x }^{ 2 }) } }{ x } } }$ 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 ${ 1 }^{ \infty }$ form :

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

- 5 years, 8 months ago

Oh thanks

- 5 years, 8 months ago

Yeah but explain.

- 5 years, 8 months ago

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

- 5 years, 8 months ago

Thanks

- 5 years, 8 months ago