Question

$\large \displaystyle \sf \red{ \lim_{ n \to \infty} {n}^{2} \int_{0}^{ \frac{1}{n} } {x}^{2018x + 1} \: dx}$​

Answer 1

Answer:

$\huge \pink{ \frac{1}{2} }$

Step-by-step explanation:

Given limit can be written as

$\displaystyle \lim_{ n \to \infty} \frac{\int_{0}^{ \frac{1}{n} } {x}^{2018x + 1} \: dx}{ \frac{1}{ {n}^{2} } }$

Its quite easy to show that it's now 0/0 form of limit . So

Now use L hospital rule . To differentiate numerator we will use fundamental theorem of calculas .

$\boxed{\red{{\frac{d}{dx} \int_{0} ^{y} f(t) \: dt = f(y). \frac{dy}{dx}} }}$

Our limit becomes

$\displaystyle \lim_{ n \to \infty} \: \frac{( \frac{1}{n} ) ^{2018. \frac{1}{n} + 1} . \frac{d}{dn}( \frac{1}{n} )}{ \frac{ - 2}{ {n}^{3} } } \\ \\ = \displaystyle \lim_{ n \to \infty} \: \frac{\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n} } \frac{1}{n} . ( \frac{ - 1}{n {}^{2} } )}{ \frac{ - 2}{ {n}^{3} } } \\ \\ = \displaystyle \lim_{ n \to \infty} \: \frac{\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n}} } {2} \\ \\ = \displaystyle \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ \log\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n}} \bigg \} \\ \\ = \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ \frac{2018}{n} \log\bigg( \frac{1}{n} \bigg) \bigg \} \\ \\ again \: l \: hopital \: rule \\ \\ = \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ 2018\bigg( \frac{1}{ \frac{1}{n} } \bigg) .\frac{ - 1}{ {n}^{2} } \bigg \} \\ \\ = \frac{1}{2} \exp \{0 \} = \frac{1}{2}$

Answer 2

Answer:

Step-by-step explanation:

The answer is 1/2 {0} 1/2