Evaluating Limit of Homework Statement

[azatkgz]

Homework Statement

Evaluate limit in terms of the number $$\alpha=\lim_{x\rightarrow 0}\frac{sinx}{x}$$


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

The Attempt at a Solution

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

 

[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'hospital's 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.

 

[azatkgz]

So the answer is just 2.

 

[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'hospital's 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.