# Evalution of a Limit

1. Oct 18, 2007

### azatkgz

1. The problem statement, all variables and given/known data
Evaluate limit in terms of the number $$\alpha=\lim_{x\rightarrow 0}\frac{sinx}{x}$$

$$\lim_{x\rightarrow 0}\frac{tan^2x+2x}{(x+x^2)}$$

3. The attempt at a solution

$$\lim_{x\rightarrow 0}\frac{tan^2x+2x}{x}-\lim_{x\rightarrow 0}\frac{tan^2x+2x}{(1+x)}$$
$$=\lim_{x\rightarrow 0}\frac{sin^2x}{xcos^2x}-\lim_{x\rightarrow 0}\frac{sin^2x}{cos^2x(1+x)}+2$$

2. Oct 18, 2007

### Kreizhn

Everything looks right to me, now just evaluate the individual limits.

Since $$\lim_{x\rightarrow 0}\frac{sin^2x}{cos^2x(1+x)}$$
is continuous about 0, then you can simply plug $$x =0$$.

For $$\lim_{x\rightarrow 0}\frac{sin^2x}{xcos^2x}$$ you can split this into

$$\lim_{x\rightarrow 0}\frac{sin(x)}{x}\lim_{x\rightarrow 0}\frac{sin(x)}{cos^2(x)}$$ or $$\lim_{x\rightarrow 0}\frac{sin^2x}{x}\lim_{x\rightarrow 0}\frac{1}{cos^2x}$$

Then use L'Hopitals rule. The answer is the same either way. (note that we've assumed here that the limits exist so that we can use the multiplicative limit law).

By the hint in the question, I assume you should do it the first way.

3. Oct 18, 2007

### azatkgz

So the answer is just 2.

4. Oct 18, 2007

### Kreizhn

That's what I got, and if you plot the function, you'll see it to be true.

Edit: If you want another way of verifying, apply L'hopitals rule right off the bat, and you'll get

$$\lim_{x\rightarrow 0}\frac{2tan(x)sec^2(x)+2}{1+2x}$$

Again a continuous function about x=0, so you can evaluate very quickly to get the same answer.

Last edited: Oct 18, 2007