My Math Forum Work check on a trig. limit

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 August 7th, 2012, 02:21 AM #1 Newbie   Joined: Sep 2011 Posts: 23 Thanks: 0 Work check on a trig. limit I am hoping someone can double check to make sure I didn't do anything illegal (mathematically), especially in step 2 and 3. I am not sure if I just made things up, or if I am allowed to work it this way by some math principles? Thanks! Robert $\quad \underset { x\rightarrow 0 }{ lim } \quad \frac { sin3x }{ { 5x }^{ 3 }-4x } \\=\underset { x\rightarrow 0 }{ lim } \quad \frac { sin3x }{ x } \cdot \quad \underset { x\rightarrow 0 }{ lim } \quad \frac { 1 }{ { 5x }^{ 2 }-4 } \\ =3\cdot \underset { x\rightarrow 0 }{ lim } \quad \frac { sinx }{ x } \cdot \quad \underset { x\rightarrow 0 }{ lim } \quad \frac { 1 }{ { 5x }^{ 2 }-4 } \\ =3\cdot 1\cdot \frac { 1 }{ -4 } \quad =\quad \frac { 3 }{ -4 }$
 August 7th, 2012, 02:27 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Work check on a trig. limit You cannot factor the 3 out of the sine function's argument (even though this coincidentally produces the correct result). I would write: $\lim_{x\to0}\frac{\sin(3x)}{5x^3-4x}=\lim_{x\to0}\frac{\sin(3x)}{3x}\cdot\lim_{x\to 0}\frac{3}{5x^2-4}=1\cdot\frac{3}{-4}=-\frac{3}{4}$
 August 7th, 2012, 02:43 AM #3 Newbie   Joined: Sep 2011 Posts: 23 Thanks: 0 Re: Work check on a trig. limit Do you know why we can not factor out the 3 if it produces the same answer? I have done three problems like this and have got the answer correct. Check this one out: $\quad \underset { t\rightarrow 0 }{ lim } \quad \frac { tan\quad 6t }{ sin\quad 2t }=\frac { \underset { t\rightarrow 0 }{ lim } \quad tan\quad 6t }{ \underset { t\rightarrow 0 }{ lim } \quad sin\quad 2t } =\frac { 6\cdot \underset { t\rightarrow 0 }{ lim } \quad tan\quad t }{ 2\cdot \underset { t\rightarrow 0 }{ lim } \quad sin\quad t } =\frac { 6\cdot \underset { t\rightarrow 0 }{ lim } \quad \frac { tan\quad t }{ t } }{ 2\cdot \underset { t\rightarrow 0 }{ lim } \quad \frac { sin\quad t }{ t } } =\frac { 6 }{ 2 } =\quad 3$ I am wondering why its wrong when I am getting them right?
 August 7th, 2012, 02:53 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Work check on a trig. limit The reason I would give is that for $k\ne0,1$, we have: $\sin$$k\theta$$\ne k\cdot\sin$$\theta$$$ The second problem, I would solve as follows (barring the use of L'Hôpital's rule): $\lim_{t\to0}\frac{\tan(6t)}{\sin(2t)}=\frac{6\lim_ {t\to0}\frac{\tan(6t)}{6t}}{2\lim_{t\to0}\frac{\si n(2t)}{2t}}=\frac{6}{2}=3$
 August 7th, 2012, 02:14 PM #5 Global Moderator   Joined: May 2007 Posts: 6,433 Thanks: 561 Re: Work check on a trig. limit (x -> 0) Lim sin(kx)/x = klim sin(kx)/kx = k. That is why it works.

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