# Thread: Can someone verify this L'Hospital problem

1. ## Can someone verify this L'Hospital problem

$\displaystyle \lim \rightarrow \infty \frac{e^{-5x}}{e^{3x}}$

$\displaystyle \lim \rightarrow \infty \frac{-5e^{-5x}}{3e^{3x}}$

$\displaystyle \lim \rightarrow \infty \frac{-5}{3e^{3x}e^{5x}}=-5$

2. Originally Posted by redsox25
$\displaystyle \lim \rightarrow \infty \frac{e^{-5x}}{e^{3x}}$

$\displaystyle \lim \rightarrow \infty \frac{-5e^{-5x}}{3e^{3x}}$

$\displaystyle \lim \rightarrow \infty \frac{-5}{3e^{3x}e^{5x}}=-5$

Any positive number raised to any power is always positive ==> the limit cannot be a negative number.

When x --> oo, e^(-3x) --> 0 and e^5x --> oo, so you can't uise L'Hospital as you did.

The limit is e^(-3x)*1/(e^(5x)) --> 0*0 = 0

Tonio

3. Originally Posted by redsox25
$\displaystyle \lim \rightarrow \infty \frac{e^{-5x}}{e^{3x}}$

$\displaystyle \lim \rightarrow \infty \frac{-5e^{-5x}}{3e^{3x}}$

$\displaystyle \lim \rightarrow \infty \frac{-5}{3e^{3x}e^{5x}}=-5$
$\displaystyle \lim_{x \to \infty} \frac{e^{-5x}}{e^{3x}}$

you can't use L'Hopital in this case.
numerator goes to $\displaystyle 0$ , denominator goes to $\displaystyle \infty$

try re-writing the expression ...

$\displaystyle \lim_{x \to \infty} \frac{1}{e^{8x}} = 0$