T Tom92 May 2010 3 0 May 26, 2010 #1 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...

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 May 26, 2010 #2 Tom92 said: 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... Click to expand... 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 \(\displaystyle \frac{1}{0}\). Reactions: Tom92

Tom92 said: 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... Click to expand... 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 \(\displaystyle \frac{1}{0}\).

S Sudharaka Dec 2009 872 381 1111 May 26, 2010 #3 Prove It said: 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 \(\displaystyle \frac{1}{0}\). Click to expand... 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 said: 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 \(\displaystyle \frac{1}{0}\). Click to expand... 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 May 26, 2010 #4 Sudharaka said: 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." Click to expand... 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 said: 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." Click to expand... 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...

S Sudharaka Dec 2009 872 381 1111 May 26, 2010 #5 Prove It said: 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... Click to expand... 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.

Prove It said: 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... Click to expand... 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.