How can I calculate $\lim\limits_{x\to\infty}x\left(\int_0^x te^{-2t}\,dt-\frac14\right)$? I tried
$$\displaystyle \int_0^x te^{-2t}\,dt$$
Let $u=t \implies du=dt$
And $dv=e^{-2t}\,dt \implies v=-\dfrac{1}{2}e^{-2t}$
$$\displaystyle \int_0^x te^{-2t}\,dt=-\dfrac{1}{2}te^{-2t}\bigg|_0^x+\int_0^x \dfrac{1}{2}e^{-2t}\,dt$$
$$=-\dfrac{1}{2}xe^{-2x}-\dfrac{1}{4}e^{-2x}+\dfrac{1}{4}$$
$$\displaystyle \lim_{x\to\infty}x\left(\int_0^x te^{-2t}\,dt-\dfrac{1}{4}\right)$$
$$=\displaystyle \lim_{x\to\infty}x\left(-\dfrac{1}{2}xe^{-2x}-\dfrac{1}{4}e^{-2x}+\dfrac{1}{4}-\dfrac{1}{4}\right)$$
$$=\displaystyle \lim_{x\to\infty}-\dfrac{1}{4}xe^{-2x}(2x+1)$$
$$=\displaystyle \lim_{x\to\infty}-\dfrac{x(2x+1)}{4e^{2x}}$$
$$=\displaystyle \lim_{x\to\infty}-\dfrac{2x^2+x}{4e^{2x}}$$
 A: Applying L’ Hospital’s rule to avoid the indeterminate form due to direct substitution, we have
$=\displaystyle \lim_{x\to\infty}-\dfrac{4x+1}{8e^{2x}}$
Indeterminate form again, applying L’ Hospital’s Rule yields
$=\displaystyle \lim_{x\to\infty}-\dfrac{4}{16e^{2x}}$
$=\displaystyle \lim_{x\to\infty}-\dfrac{1}{4e^{2x}}$
$=0$
Hence
$\displaystyle \lim_{x\to\infty}x\left(\int_0^x te^{-2t}\,dt-\dfrac{1}{4}\right)=0$
A: You can  avoid computing the integral by writing
$$\lim_{x\to \infty}x\left (\int_0^xte^{-2t}dt-\frac{1}{4}\right)=\lim_{x\to \infty}\frac{\int_0^xte^{-2t}dt-\frac{1}{4}}{\frac 1x}$$
Applying L'Hospital's rule, if the limit exists the second limit is equal to
$$\lim_{x\to \infty}\frac{xe^{-2x}}{-\frac{1}{x^2}}=\lim_{x\to \infty}-\frac{x^3}{e^{2x}}$$
which is $0$ by using L'Hospital three times.
A: $$\int_{0}^{x}t e^{-2t}\,dt -\frac{1}{4} = -\int_{x}^{+\infty}t e^{-2t}\,dt=-e^{-2x}\int_{0}^{+\infty}(t+x)e^{-2t}\,dt $$
and
$$ 0\leq \int_{0}^{+\infty}(t+x)e^{-2t}\,dt = x^2\int_{0}^{+\infty}(t+1)e^{-2t x}\,dt\leq Cx^2 $$
hence the limit is trivially zero by squeezing.
