# Limits using L'Hospital's rule.

• Nov 6th 2010, 05:15 PM
ioke09
Limits using L'Hospital's rule.
Evaluate the limit using L'Hospital's rule.

y = lim x->infinity (5x-1/5x+3)^5x+2

$ln y = (5x + 2) ln (5x-1/5x+3)$

$lim x-> infinity$ lny= lim x-> infinity ln (5x-1)-ln (5x+3)/ 1/(5x+2)= lim x-> infinity 5/(5x − 1) − 5/(5x + 3)/−5/(5x + 2)^2

I'm not sure how to proceed after this step.

Any help would be appreciated, thank you for your time and effort :D
• Nov 6th 2010, 07:17 PM
mr fantastic
Quote:

Originally Posted by ioke09
Evaluate the limit using L'Hospital's rule.

y = lim x->infinity (5x-1/5x+3)^5x+2

$ln y = (5x + 2) ln (5x-1/5x+3)$

$lim x-> infinity$ lny= lim x-> infinity ln (5x-1)-ln (5x+3)/ 1/(5x+2)= lim x-> infinity 5/(5x − 1) − 5/(5x + 3)/−5/(5x + 2)^2

I'm not sure how to proceed after this step.

Any help would be appreciated, thank you for your time and effort :D

Note that $\displaystyle \frac{5x - 1}{5x + 3} = 1 - \frac{4}{5x + 3}$. Therefore l'Hospital's Rule is not required if you're familiar with the basic limit $\displaystyle \lim_{t \to +\infty} \left(1 + \frac{a}{t} \right)^t = e^a$.

In your limit, substitute $t = 5x + 3$, re-arrange using a basic index law and then use the limit of a product theorem. It becomes:

$\displaystyle \lim_{t \to +\infty} \left[\left(1 - \frac{4}{t} \right)^{-1} \left(1 - \frac{4}{t} \right)^t\right] = e^{-4}$.