# Thread: Limits using L'Hospital's rule.

1. ## Limits using L'Hospital's rule.

Evaluate the limit using L'Hospital's rule.

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

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

$\displaystyle lim x-> infinity$ 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

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

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

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

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

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

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