# Limit involving trigonometry

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1. Oct 17, 2014

### Sheepwall

1. The problem statement, all variables and given/known data
"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."

2. Relevant equations
$(1) \sin(a\pm b) = \cos(a)\sin(b)\pm\cos(b)\sin(a)$

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

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

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

3. 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.

Last edited: Oct 17, 2014
2. Oct 17, 2014

### LCKurtz

Try L'Hospital's rule.

3. Oct 17, 2014

### Sheepwall

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

4. Oct 17, 2014

### 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

5. Oct 17, 2014

### Sheepwall

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

6. Oct 17, 2014

### ehild

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

7. Oct 18, 2014

### Sheepwall

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

8. Oct 18, 2014

Clever! :)