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what is the lime of $ \frac{1-e^{-kt}}{k}$ as $k \to \infty$?

Is that just equal $\frac{1}{\infty}=0$?

Does any one can help, I am not sure if We can apply L'Hopital's rule. S

user58228
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    Why do you need L'Hopital's rule? denominator goes to infinity and numerator is bounded when $t\geq 0$, your estimation is correct. When $t<0$, you need the rule. – newbie May 28 '13 at 14:38

1 Answers1

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HINT:

If $t=0,e^{-kt}=1$

If $t>0, \lim_{k\to\infty}e^{-kt}=0$

If $t<0, t=-r^2$(say), $\lim_{k\to\infty}\frac{1-e^{-kt}}k=\lim_{k\to\infty}\frac{1-e^{kr^2}}k=\frac\infty\infty $

So, applying L'Hospitals' rule, $\lim_{k\to\infty}\frac{1-e^{kr^2}}k=-r^2\cdot\lim_{k\to\infty}\frac{e^{r^2t}}1=-r^2\cdot\infty=-\infty$

lab bhattacharjee
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