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I tried

$$\displaystyle \int_0^x te^{-2t}\,dt$$

Let $u=t \implies du=dt$ And $dv=e^{-2t}\,dt \implies v=-\dfrac{1}{2}e^{-2t}$

$$\displaystyle \int_0^x te^{-2t}\,dt=-\dfrac{1}{2}te^{-2t}\bigg|_0^x+\int_0^x \dfrac{1}{2}e^{-2t}\,dt$$ $$=-\dfrac{1}{2}xe^{-2x}-\dfrac{1}{4}e^{-2x}+\dfrac{1}{4}$$ $$\displaystyle \lim_{x\to\infty}x\left(\int_0^x te^{-2t}\,dt-\dfrac{1}{4}\right)$$ $$=\displaystyle \lim_{x\to\infty}x\left(-\dfrac{1}{2}xe^{-2x}-\dfrac{1}{4}e^{-2x}+\dfrac{1}{4}-\dfrac{1}{4}\right)$$ $$=\displaystyle \lim_{x\to\infty}-\dfrac{1}{4}xe^{-2x}(2x+1)$$ $$=\displaystyle \lim_{x\to\infty}-\dfrac{x(2x+1)}{4e^{2x}}$$ $$=\displaystyle \lim_{x\to\infty}-\dfrac{2x^2+x}{4e^{2x}}$$

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    apply L’ Hospital’s rule to avoid the indeterminate form. – 2017-01-07
  • 1
    Notice that you could apply L'Hospital's rule (twice) from the *very beginning*. – 2017-01-07

3 Answers 3

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

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    A neat answer (+1). – 2017-01-07
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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.

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