1. ## L'Hospitals Rule Troubles

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.

2. Originally Posted by sfgiants13
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:
$\lim _{t = \to \infty } \left[ {\frac{{ - t}}
{{5e^{5t} }} - \frac{1}
{{25e^{5t} }}} \right] = \lim _{t = \to \infty } \left[ {\frac{{ - 5t - 1}}
{{25e^{5t} }}} \right] = \frac{{ - 1}}
{{25}}$
Is that really what you meant?

3. Originally Posted by Plato
Here is what you have posted:
$\lim _{t = \to \infty } \left[ {\frac{{ - t}}
{{5e^{5t} }} - \frac{1}
{{25e^{5t} }}} \right] = \lim _{t = \to \infty } \left[ {\frac{{ - 5t - 1}}
{{25e^{5t} }}} \right] = \frac{{ - 1}}
{{25}}$
Is that really what you meant?

I don't know how to use the math tags but the equatino above isn't right.

(-1/5) x te^-5t - (1/25) x e^-5t as t approaches infinity.

4. Originally Posted by sfgiants13
the equatino above isn't right.
(-1/5) x te^-5t - (1/25) x e^-5t as t approaches infinity.
Well it surely is the same to me.
$\left[ {\left( { - 1/5} \right)\left( {te^{ - 5t} } \right) = \frac{{ - t}}
{{5e^t }}\,} \right]\& \,\left[ { - \left( {1/25} \right)\left( {e^{ - 5t} } \right) = \frac{{ - 1}}{{25e^{5t} }}} \right]$

What is incorrect about the above?

5. I got -1/25 too for some reason but the book says it's convergant to 1/25. I'm thinking ti was just a math error in the book now...

6. Originally Posted by sfgiants13
I got -1/25 too for some reason but the book says it's convergant to 1/25. I'm thinking ti was just a math error in the book now...
$\left[ {\left( { {\color{red}-} 1/5} \right)\left( {te^{ - 5t} } \right) = \frac{{ {\color{red}-} t}}{{5e^t }}\,} \right]$
I think that there is indeed a typo but I think it is in that first minus sign.