Trigonometric Limit without L'Hôpital's Rule

[Sheepwall]

Homework Statement

"Calculate the following limit if it exists. If it does not exist, motivate why.
$\displaystyle\lim_{x\rightarrow 0} {\frac{x + x^2 +\sin(3x)}{tan(2x) + 3x}}$

Do not use l'Hôpital's rule."

Homework Equations

$(1) \sin(a\pm b) = \cos(a)\sin(b)\pm\cos(b)\sin(a)$

$(2) \cos(a\pm b) = \cos(a)\cos(b)\mp\sin(a)\sin(b)$

$(3) \displaystyle\lim_{x\rightarrow 0} {\frac{\sin(x)}{x}} = 1$

$(4) \tan(x) = \frac{\sin(x)}{\cos(x)}$

The Attempt at a Solution

I have tried expressing the trigonometrics in terms of $\sin(x)$ and $\cos(x)$, but it just got messier without helping me in any way.

This isn't me just jumping on these forums as soon as I can't find the answer; I have genuinely been trying to solve this problem and looking over my methods much more than once.

Thanks in advance!

 

[LCKurtz]

Try L'Hospital's rule.

 

[Sheepwall]

Ah! Forgot to mention: L'Hôpital's rule is prohibited on this exercise. Sorry, I'll add it to the post.

 

[ehild]

Divide both the numerator and denominator by x. Use that the limit of sin(kx)/(kx) is zero if x goes to zero and k is a constant.

ehild

 

[Sheepwall]

Thanks, I'll try that. Don't you mean the limit $\displaystyle\lim_{x\rightarrow 0} {\frac{\sin(x)}{x}} = 1$ though?

 

[ehild]

Thanks, I'll try that. Don't you mean the limit $\displaystyle\lim_{x\rightarrow 0} {\frac{\sin(x)}{x}} = 1$ though?

Yes, I meant that, but you have sin3x and sin(2x) so consider the limit of sin(kx)/x .

 

[Sheepwall]

Thank you for the help, I solved it yesterday by dividing numerator and denominator by 3x and realizing that 3x = 2x * 1.5.

 

[ehild]

Thank you for the help, I solved it yesterday by dividing numerator and denominator by 3x and realizing that 3x = 2x * 1.5.

Clever! :)