Limit Exercise asked by Dan S in many Math groups Level 3

Exercise:

Find $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\int_{1}^{x}{{{t}^{t-1}}\left( t+t\ln t+1 \right)dt}}{{{x}^{x}}}$

Solution: Let $f\left( x \right)=\int_{1}^{x}{{{t}^{t-1}}\left( t+t\ln t+1 \right)dt}\,\,\And \,\,g\left( x \right)={{x}^{x}}$

So $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\int_{1}^{x}{{{t}^{t-1}}\left( t+t\ln t+1 \right)dt}}{{{x}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\frac{f\left( \infty  \right)}{g\left( \infty  \right)}=\frac{\infty }{\infty }\,ind\,form$

Using L’Hopital Rule and Fundamental theorem of calculus we get :

$\frac{d}{dx}f\left( x \right)=\frac{d}{dx}\left( \int_{1}^{x}{{{t}^{t-1}}\left( t+t\ln t+1 \right)dt} \right)={{x}^{x-1}}\left( x+x\ln x+1 \right)$ and $\frac{d}{dx}g\left( x \right)=\left( 1+\ln x \right){{x}^{x}}$

So $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\int_{1}^{x}{{{t}^{t-1}}\left( t+t\ln t+1 \right)dt}}{{{x}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\frac{d}{dx}f\left( x \right)}{\frac{d}{dx}g\left( x \right)}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{x-1}}\left( x+x\ln x+1 \right)}{\left( 1+\ln x \right){{x}^{x}}}$

$=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{x-1}}{{x}^{-x}}\left( x+x\ln x+1 \right)}{\ln x+1}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{-1}}\left( x\left( 1+\ln x \right)+1 \right)}{1+\ln x}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{1+\ln x+{{x}^{-1}}}{1+\ln x}$

$=1+\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{-1}}}{1+\ln x}=1+\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{x\left( 1+\ln x \right)}=1+\frac{1}{\infty }=0+1=1$