# Thread: Evaluate the limit

1. ## Evaluate the limit

I know that i have to use L'Hos rule and to take a derevative, but I cant get it.

2. Originally Posted by kashifzaidi

I know that i have to use L'Hos rule and to take a derevative, but I cant get it.
Let $\displaystyle y=\lim_{x\to\infty}\left(\frac{5x-1}{5x+3}\right)^{5x+2}$

Take the ln of both sides and see where that gets ya.

3. Originally Posted by kashifzaidi

I know that i have to use L'Hos rule and to take a derevative, but I cant get it.
DO you have to use L'Hospital?

Note that $\displaystyle \frac{5x - 1}{5x + 3} = 1 - \frac{4}{5x + 3}$.

So $\displaystyle \left(\frac{5x - 1}{5x + 3}\right)^{5x + 2} = \left(1 - \frac{4}{5x + 3}\right)^{5x + 2}$

The stuff inside the brackets $\displaystyle \to 1$ as $\displaystyle x \to \infty$.

What does $\displaystyle 1^{5x + 2} \to$ as $\displaystyle x \to \infty$?

4. Originally Posted by kashifzaidi

I know that i have to use L'Hos rule and to take a derevative, but I cant get it.
If you substitute $\displaystyle u = 5x + 3$ the limit can be written as $\displaystyle \lim_{u \to + \infty} \left(1 - \frac{4}{u} \right)^{u - 1}$ and you can now use a well known limit involving our good friend $\displaystyle e$ to easily get the answer.

5. what is the other side ?

6. Originally Posted by kashifzaidi
what is the other side ?
y

7. I did it...but I know that answer is not from the following: 0 , INFINITY or -INFINITY

8. Originally Posted by kashifzaidi
I did it...but I know that answer is not from the following: 0 , INFINITY or -INFINITY
read this thread of related interest: http://www.mathhelpforum.com/math-he...te-limits.html