Limit of a long trig function.

[mrjoe2]

Homework Statement

lim
x->0 ((sin5x)/(sin2x) - (sin3x)/(4x))

Homework Equations

i guess sinx/x as x approaches zero is 1.

The Attempt at a Solution

lets be honest, this is a simple question. i am getting a final answer of 4, but it is wrong apperently according to the book's solution. i want to see if anyone else got this answer

I separated the limit into two limits, divided by 5x's and 2x's on the left limit and divided by 3x's on the right limit. this shouldn't be a problem (and yes i divided top and bottom by the same amounts so they would cancel out; that's not my mistake). please someone help

 

[gabbagabbahey]

You've obviously made an algebra error somewhere, If you show your work, we can tell you where your error is.

 

[Mark44]

Are you sure you have given us the problem as it appears in your book or assignment? Could it be lim sin(5x)/(2x) - sin(3x)/(4x)? If so, that's a much simpler problem, whose limit is 1.75, as x approaches 0.

 

[HallsofIvy]

\frac{sin(3x)}{4x}= \frac{3}{4}\frac{sin(3x)}{3x}

\frac{sin(5x)}{sin(2x)}= \frac{5}{2}\frac{sin(5x)}{5x}\frac{sin(2x)}{2x}

 

[mrjoe2]

Are you sure you have given us the problem as it appears in your book or assignment? Could it be lim sin(5x)/(2x) - sin(3x)/(4x)? If so, that's a much simpler problem, whose limit is 1.75, as x approaches 0.

HOW IS IT 1.75! wheres the flaw in my logic.

(sin5x)/sin2x = [((sin5x)/(5x*2x)]/[(sin2x)/(5x*2x)]
= (5/2x)/(2/5x))
= 25/4


(sin3x)/4x = [(sin3x/3x)]/[(4x/3x)]
= 3/(4x/3x)
= 3/(4/3)
= 9/4

25/4 - 9/4 = 16/4 = 4 where is my flaw!?

 

[Mark44]

\frac{sin(3x)}{4x}= \frac{3}{4}\frac{sin(3x)}{3x}

\frac{sin(5x)}{sin(2x)}= \frac{5}{2}\frac{sin(5x)}{5x}\frac{sin(2x)}{2x}

Halls, are you missing a division symbol in your second equation?

 

[Mark44]

HOW IS IT 1.75! wheres the flaw in my logic.

(sin5x)/sin2x = [((sin5x)/(5x*2x)]/[(sin2x)/(5x*2x)]
= (5/2x)/(2/5x))
= 25/4

How did you get (5/2x)/(2/5x)) from the expression above it? lim (as x \rightarrow 0) of sin(5x) / (5x) is 1, not 5, if that's what you did.

The expression on the right side of your first line is equal to
\frac{sin(5x)}{5x} * \frac{5x}{2x} * \frac{2x}{sin(2x)}
As x \rightarrow 0, this expression approaches 1 * 5/2 * 1 = 5/2. Similarly sin(3x)/(4x) approaches 3/4, so the difference approaches 7/4 = 1.75.

Regarding my earlier question about whether the denominator of the first fraction should have been 2x rather than sin(2x), it doesn't matter. For x close to 0, sin(x) \approx x, and sin(2x) \approx 2x.

(sin3x)/4x = [(sin3x/3x)]/[(4x/3x)]
= 3/(4x/3x)
= 3/(4/3)
= 9/4

25/4 - 9/4 = 16/4 = 4 where is my flaw!?