Question

Find the value of $\underset{x\to 0}{lim}{\left({\int }_0^1\right(by+a(1-y){)}^{x}dy)}^{1/x}$

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (A), (C)

The correct options are
A $1$
C $0$

Given : $\underset{x\to 0}{lim}{\left[{\int }_0^1(by+a(1-y){)}^{x}y\right]}^{\frac{1}{x}}$

$I=\underset{x\to 0}{lim}{\left[{\int }_0^1(by+a(1-y){)}^{x}y\right]}^{\frac{1}{x}}I=\underset{x\to 0}{lim}{\left[{\int }_0^1\left(\right(b-a)y+a{)}^{x}y\right]}^{\frac{1}{x}}I=\underset{x\to 0}{lim}{\left[{\left[\frac{\left(\right(b-a)y+a{)}^{x+1}}{(b-a)(x+1)}\right]}_0^1\right]}^{\frac{1}{x}}I=\underset{x\to 0}{lim}{\left[\frac{b^{(x+1)}-a^{x+1}}{(b-a)(x+1)}\right]}^{\frac{1}{x}}\left(1^{\infty }from\right)I=\underset{x\to 0}{lim}[\frac{b^{(x+1)}-a^{(x+1)}-(b-a)(x+1)}{(b-a)(x+1)}\times \frac{1}{x}]I=\underset{x\to 0}{lim}[\frac{b(b^x-1)}{x}-\frac{a(a^x-1)}{x}]\frac{1}{(b-a)}I=e\left[\frac{b\ln b-a\ln a}{(b-a)}\right]={\left[\frac{b^b}{a^a}\right]}^{\frac{1}{(b-a)}}$