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! :)