mathop {Lim}limits_{lambda to 0} ,{left( {in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $L = \mathop {Lim}\limits_{\lambda  	o 0} \,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } ight)^{\frac{1}{\lambda }}}$First,evaluate

Vedclass · Accepted Answer

$\frac{4}{e}$

Vedclass · Answer

(B) Let $L = \mathop {Lim}\limits_{\lambda  	o 0} \,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } ight)^{\frac{1}{\lambda }}}$First,evaluate the integral: $\int\limits_0^1 {{{(1 + x)}^\lambda }dx} = \left[ \frac{{{{(1 + x)}^{\lambda  + 1}}}}{{\lambda  + 1}} ight]_0^1 = \frac{{{2^{\lambda  + 1}} - 1}}{{\lambda  + 1}}$Now,$L = \mathop {Lim}\limits_{\lambda  	o 0} {\left( {\frac{{{2^{\lambda  + 1}} - 1}}{{\lambda  + 1}}} ight)^{\frac{1}{\lambda }}}$This is in the form $1^{\infty}$. We use the formula $\mathop {Lim}\limits_{x 	o a} f(x)^{g(x)} = e^{\mathop {Lim}\limits_{x 	o a} g(x)(f(x) - 1)}$:$L = e^{\mathop {Lim}\limits_{\lambda  	o 0} \frac{1}{\lambda } \left( \frac{{{2^{\lambda  + 1}} - 1}}{{\lambda  + 1}} - 1 ight)} = e^{\mathop {Lim}\limits_{\lambda  	o 0} \frac{1}{\lambda } \left( \frac{{{2^{\lambda  + 1}} - 1 - \lambda  - 1}}{{\lambda  + 1}} ight)}$$L = e^{\mathop {Lim}\limits_{\lambda  	o 0} \frac{{{2^{\lambda  + 1}} - 2 - \lambda }}{{\lambda (\lambda  + 1)}}} = e^{\mathop {Lim}\limits_{\lambda  	o 0} \left( \frac{{2({2^\lambda } - 1)}}{\lambda } - 1 ight) \cdot \frac{1}{{\lambda  + 1}}}$Since $\mathop {Lim}\limits_{\lambda  	o 0} \frac{{{2^\lambda } - 1}}{\lambda } = \ln 2$,we have:$L = e^{2 \ln 2 - 1} = e^{\ln 4 - \ln e} = e^{\ln(4/e)} = \frac{4}{e}$

$\mathop {Lim}\limits_{\lambda \to 0} \,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } \right)^{\frac{1}{\lambda }}}$ is equal to

Similar Questions

The equation of a tangent to the curve $y \cot x = y^3 \tan x$ at the point where the abscissa is $\frac{\pi}{4}$ is:

If $F(x) = \left(f\left(\frac{x}{2}\right)\right)^2 + \left(g\left(\frac{x}{2}\right)\right)^2$,where $f^{\prime \prime}(x) = -f(x)$ and $g(x) = f^{\prime}(x)$,and given $F(5) = 5$,then $F(10)$ is equal to

If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$, then $f' \left( \frac{\pi}{4} \right)$ is equal to

Let $[x]$ denote the greatest integer function,and let $m$ and $n$ respectively be the numbers of the points,where the function $f(x) = [x] + |x - 2|$,$-2 < x < 3$,is not continuous and not differentiable. Then $m + n$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module