L'Hopital's rule

May 2010
3
0
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
Aug 2008
12,883
4,999
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 \(\displaystyle \frac{1}{0}\).
 
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Dec 2009
872
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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
Aug 2008
12,883
4,999
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...
 
Dec 2009
872
381
1111
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.