# 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
3 years, 10 months ago

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

- 3 years, 10 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 :

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

- 3 years, 10 months ago

Oh thanks

- 3 years, 10 months ago

Yeah but explain.

- 3 years, 10 months ago

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

- 3 years, 10 months ago

Thanks

- 3 years, 10 months ago