# Math Help - L'Hopital's rule

1. ## L'Hopital's rule

Hi,
Why can't i use L'Hopital's rule in this limit:

lim (2x+sin2x+1)/(2x+sin2x)(sinx+3)^2 as x->infinity

Thanks...

2. Originally Posted by Tom92
Hi,
Why can't i use L'Hopital's rule in this limit:

lim (2x+sin2x+1)/(2x+sin2x)(sinx+3)^2 as x->infinity

Thanks...
I assume this is

$\lim_{x \to \infty}\frac{2x + \sin{2x} + 1}{(2x + \sin{2x})(\sin{x} + 3)^2}$.

You can only use L'Hospital's Rule if you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ from direct substitution.

But in this case, you'll get $\frac{1}{0}$.

3. Originally Posted by Prove It
I assume this is

$\lim_{x \to \infty}\frac{2x + \sin{2x} + 1}{(2x + \sin{2x})(\sin{x} + 3)^2}$.

You can only use L'Hospital's Rule if you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ from direct substitution.

But in this case, you'll get $\frac{1}{0}$.
Dear Prove It,

I think you have made mistake in this case. Note that the limit goes to "infinity". Hence the denominator and numerator both tends to "infinity."

4. Originally Posted by Sudharaka
Dear Prove It,

I think you have made mistake in this case. Note that the limit goes to "infinity". Hence the denominator and numerator both tends to "infinity."
Oh, oops. Hahaha, I subsituted $x = 0$ instead.

Then I don't see why you can't use L'Hospital's Rule in this case...

5. Originally Posted by Prove It
Oh, oops. Hahaha, I subsituted $x = 0$ instead.

Then I don't see why you can't use L'Hospital's Rule in this case...
Dear Prove It,

But notice that,

$\lim_{x\rightarrow{\infty}}\frac{2x+\sin{2x}+1}{(2 x+\sin{2x})(\sin{x}+3)^2}$

Dividing the numerator and denominator by "2x";

$\lim_{x\rightarrow{\infty}}\frac{1+\frac{\sin{2x}} {2x}+\frac{1}{2x}}{(1+\frac{\sin{2x}}{2x})(sinx+3) ^2}$

Therefore the limit does not exist.