Calculate and evaluate the limit:
$$\lim_{x\to {\infty}} \frac{\int_1^x (t^2(e^{1/t}-1)-t)\,dt}{x^2\ln\left(1+\frac{1}{x}\right)}$$
When plotting the upper and the lower part of the fraction separately it becomes clear that it is a $\frac{\infty}{\infty}$ case. However, I can't solve for the integral. Also, it is not totally clear to me why the limit on the lower part approaches ${\infty}$ and not ${0}$ (considering it approaches ${{{\infty}^2}{\ln(1)}}$). Thank you a lot to everyone that can helps me with it somehow.
Why lower part approaches zero: $$x^2ln(1+1/x)=\frac{ln(1+1/x)}{1/x^2}$$which is of the form $0/0$. You can use L'Hospital's Rule to show: $$\lim_{x\to \infty}\frac{ln(1+1/x)}{1/x^2}=\lim_{x\to \infty}\frac{x}{1+1/x}$$ which goes to infinity.
Now, let $$L=\lim_{x\to {\infty}} \frac{\int_1^x (t^2(e^{1/t}-1)-t)\,dt}{x^2\ln(1+1/x)}$$ Again, L is of the form $\infty/\infty$. Again, use L'Hospital's Rule: $$L=\frac{x^2(e^{1/x}-1)-x}{2x\text{ln}(1+1/x)-\frac{1}{1+1/x}}$$ Now as mentioned in one of the comments, you can use various expansions to obtain: $$L=\lim_{x\to \infty}\frac{1/2+1/3x+\cdots}{1-1/3x^2+\cdots}$$ $$L=\frac{1}{2}$$ I might have done some calculation mistake in the very last part, but I hope you know how to handle such problems now...