Why does the limit as t approaches infinity of [(-te-5t)/5]-[e^-5t/25]= 1/25? I keep getting infinities when I try to use l'Hospitals rule.
Here is what you have posted:
$\displaystyle \lim _{t = \to \infty } \left[ {\frac{{ - t}}
{{5e^{5t} }} - \frac{1}
{{25e^{5t} }}} \right] = \lim _{t = \to \infty } \left[ {\frac{{ - 5t - 1}}
{{25e^{5t} }}} \right] = \frac{{ - 1}}
{{25}}$ Is that really what you meant?
Well it surely is the same to me.
$\displaystyle \left[ {\left( { - 1/5} \right)\left( {te^{ - 5t} } \right) = \frac{{ - t}}
{{5e^t }}\,} \right]\& \,\left[ { - \left( {1/25} \right)\left( {e^{ - 5t} } \right) = \frac{{ - 1}}{{25e^{5t} }}} \right]$
What is incorrect about the above?