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.