Mathematics - Definite Integration Question with Solution | TestHub

For $l_1$ : using L Hospital rule and Leibnitz's rule, we get\ Given limit $={\lim }_{x\to 0}\left(\frac{x\left(e^{3x}-1\right)\ln (1+2x)}{3x^2\left(x^3+3\right)(1-\cos \sqrt{x})}\right)$ $={\lim }_{x\to 0}\frac{\left(\frac{e^3-1}{3x}\right)\frac{\ln (1+2x)}{2x}\times 3x\times 2x}{3x\left(x^3+3\right)\frac{(1-\cos \sqrt{x})}{(\sqrt{x}{)}^2}(\sqrt{x}{)}^2}\cdot =\frac{(1)(1)(2)}{(3)\left(\frac{1}{2}\right)}=\frac{4}{3}$.

So, $l_1=\frac{4}{3}$.