how to evalute trigonometric limits

EriLynn.

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Oct 4, 2009
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So i'm just reviewing for a course i will be taking soon, I took a calculus last year, and for some reason i cannot recall, nor find in my text book, how to solve these kinds of questions.
A question like " limit as x approaches 0 of (sin(2x)/3x) ....
I tried changing the sin(2x) into 2sinxcosx... but after that i dont know what to do
could anyone help me out, i just need to know the idea and what kind of steps to follow
thanks a million folks!
 
You do not have to change sin(2x) to 2sin(x)cos(x).

Do this, bring out the 1/3 outside the limit:

13limx0sin(2x)x\displaystyle \frac{1}{3}\lim_{x\to 0}\frac{sin(2x)}{x}

Multiply top and bottom by 2:

13limx02sin(2x)2x\displaystyle \frac{1}{3}\lim_{x\to 0}\frac{2sin(2x)}{2x}

23limx0sin(2x)2xlimit is 1\displaystyle \frac{2}{3}\underbrace{\lim_{x\to 0}\frac{sin(2x)}{2x}}_{\text{limit is 1}}

231=23\displaystyle \frac{2}{3}\cdot 1=\frac{2}{3}
 
Another way:

limx0sin(2x)3x gives the indeterminate form 00.\displaystyle \lim_{x\to0}\frac{sin(2x)}{3x} \ gives \ the \ indeterminate \ form \ \frac{0}{0}.

Ergo, the Marqui to the rescue. Dx[sin(2x)]Dx(3x) = 2cos(2x)3,\displaystyle Ergo, \ the \ Marqui \ to \ the \ rescue. \ \frac{D_x[sin(2x)]}{D_x(3x)} \ = \ \frac{2cos(2x)}{3},

hence limx02cos(2x)3 = 2(1)3 = 23.\displaystyle hence \ \lim_{x\to0}\frac{2cos(2x)}{3} \ = \ \frac{2(1)}{3} \ = \ \frac{2}{3}.
 
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