Waste less time on Facebook — follow Brilliant.
×

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
2 years, 4 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

You can use the fact that \( \displaystyle \lim_{x \rightarrow 0} f(x)^{g(x)} = \displaystyle \lim_{x \rightarrow 0} e^{g(x)(f(x) - 1)} \) Siddhartha Srivastava · 2 years, 4 months ago

Log in to reply

@Siddhartha Srivastava Thanks Gautam Sharma · 2 years, 4 months ago

Log in to reply

Answer Should be \(\lambda =3\) . It is \({ 1 }^{ \infty }\) form. Deepanshu Gupta · 2 years, 4 months ago

Log in to reply

@Deepanshu Gupta Yeah but explain. Gautam Sharma · 2 years, 4 months ago

Log in to reply

@Deepanshu Gupta 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 :

\(L=\lim _{ x\rightarrow a }{ { f\left( x \right) }^{ g\left( x \right) } } ={ e }^{ \lim _{ x\rightarrow a }{ { (f\left( x \right) -1)g\left( x \right) } } }\\ here:\quad \lim _{ x\rightarrow a }{ { f\left( x \right) } } =1\quad \& \quad \lim _{ x\rightarrow a }{ { g\left( x \right) } } =\infty \quad \) Deepanshu Gupta · 2 years, 4 months ago

Log in to reply

@Deepanshu Gupta Oh thanks Gautam Sharma · 2 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...