Mathematics - Limits Question with Solution | TestHub

Consider ${\lim }_{x\to \infty }\frac{{\int }_0^{2x}x{e}^{x^2}dx}{e^{4x^2}}$ $$\begin{array}{l}={\lim }_{x\to \infty }\frac{2{\int }_0^{2x}x{e}^{x^2}dx}{2e^{4x^2}}\\ ={\lim }_{x\to \infty }\frac{2{\int }_0^{2x}{e}^{x^2}d\left(x^2\right)}{2e^{4x^2}}\\ ={\lim }_{x\to \infty }\frac{{\left[e^{x^2}\right]}_0^{2x}}{e^{4x^2}}={\lim }_{x\to \infty }\frac{e^{4x^2}-1}{2e^{4x^2}}\\ ={\lim }_{x\to \infty }\left(\frac{1}{2}-\frac{1}{2e^{4x^2}}\right)=\frac{1}{2}\end{array}$$