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 $y=\lim_{x\to\infty}\left(\frac{5x-1}{5x+3}\right)^{5x+2}$

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

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

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

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

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 $u = 5x + 3$ the limit can be written as $\lim_{u \to + \infty} \left(1 - \frac{4}{u} \right)^{u - 1}$ and you can now use a well known limit involving our good friend $e$ 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