The Limit of a Trigonometric Function

[Jambr]

Homework Statement

lim x-> 0 ( sin(5x) - 3x ) / ( 3x )

Homework Equations

(a - b) / c = (a / c) - b

The Attempt at a Solution

(1/3) lim x->0 ( sin(5x) / x ) - 3x
= (1/3) lim x->0 ( sin(5x) / x )( 5 / 5) - 3x
= (5/3) lim x->0 ( sin(5x) / 5x ) - 3x
= (5/3) * 1 - 3x
= (5/3) - 3x
= (5/3)

I use WolframAlpha to check whether or not my answers are correct. WolframAlpha and I are having a disagreement (even though I am most likely wrong) about this problem. I say 5/3, it says 2/3. Can someone please explain where I went wrong?

Sorry, I tried using LaTeX but I just got annoyed.

 

[Dick]

You can start with (a - b) / c = (a / c) - b. That's not right. (a-b)/c=(a/c)-(b/c).

 

[HallsofIvy]

Homework Statement

lim x-> 0 ( sin(5x) - 3x ) / ( 3x )

Homework Equations

(a - b) / c = (a / c) - b

The Attempt at a Solution

(1/3) lim x->0 ( sin(5x) / x ) - 3x
= (1/3) lim x->0 ( sin(5x) / x )( 5 / 5) - 3

Here is your error.
\frac{sin(5x)- 3x}{3x}= \frac{sin(5x)}{3x}-\frac{3x}{3x}= \frac{sin(5x)}{3x}- 1
= \frac{5}{3}\frac{sin(5x)}{5x}- 1

= (5/3) lim x->0 ( sin(5x) / 5x ) - 3x
= (5/3) * 1 - 3x
= (5/3) - 3x
= (5/3)

I use WolframAlpha to check whether or not my answers are correct. WolframAlpha and I are having a disagreement (even though I am most likely wrong) about this problem. I say 5/3, it says 2/3. Can someone please explain where I went wrong?

Sorry, I tried using LaTeX but I just got annoyed.

 

[Jambr]

Thank you, I had tried that already but I did it by removing the 3 from the bottom first off, which was wrong.

Another question: As we all know sin x / x = 1, is it also true at sin^2 x / x^2 = 1?

 

[Dick]

Thank you, I had tried that already but I did it by removing the 3 from the bottom first off, which was wrong.

Another question: As we all know sin x / x = 1, is it also true at sin^2 x / x^2 = 1?

If you mean limit as x->0, sure. Both limits are 1.