How can I calculate $\lim\limits_{x\to\infty}x\left(\int_0^x te^{-2t}\,dt-\frac14\right)$?

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 $\displaystyle \lim_{x\to\infty}x\left(\int_0^x te^{-2t}\,dt-\dfrac{1}{4}\right)=0$

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.

$$\int_{0}^{x}t e^{-2t}\,dt -\frac{1}{4} = -\int_{x}^{+\infty}t e^{-2t}\,dt=-e^{-2x}\int_{0}^{+\infty}(t+x)e^{-2t}\,dt $$ 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.