# L'Hopital's rule

#### 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...

#### Prove It

MHF Helper
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

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

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

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

Tom92

#### Sudharaka

I assume this is

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

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

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

#### Prove It

MHF Helper
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 $$\displaystyle x = 0$$ instead.

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

#### Sudharaka

Oh, oops. Hahaha, I subsituted $$\displaystyle x = 0$$ instead.

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

But notice that,

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

Dividing the numerator and denominator by "2x";

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

Therefore the limit does not exist.