Limits and L'Hospital's Rule

1. L'Hospital's Rule 1 ${\textstyle \lim _{x\to 0}{f(x) \over g(x)}=lim_{x\to 0}{f'(x) \over g'(x)}}$
2. L'Hospital's Rule 2 ${\textstyle \lim _{x\to \infty }{f(x) \over g(x)}=lim_{x\to \infty }{f'(x) \over g'(x)}}$

there are some of the problems that can't use the L'Hospital's Rule such as:

${\textstyle \lim _{x\to \infty }{x+cosx \over x}}$

If you are only using the L'Hospital's Rule this is what you will get:

Start ${\textstyle \lim _{x\to \infty }{x+cosx \over x}}$

1. Derivatives${\textstyle \lim _{x\to \infty }{1-sinx \over 1}}$
2. Derivatives${\textstyle \lim _{x\to \infty }{-cosx}}$
3. Derivatives${\textstyle \lim _{x\to \infty }{sinx}}$

And then so on and forth so that means the limit does not exist.

So if you wan't to compute this first limit we can't use L'Hospital's Rule and this is how we do it.

Start ${\textstyle \lim _{x\to \infty }{x+cosx \over x}}$

1. ${\textstyle \lim _{x\to \infty }{1+{cos \over x}}}$

And then we will get the answer 1. because cos x only can be between 1 to -1 but it has been divided by Infinite so we can say it is almost zero and then + 1 so we will get 1.