I need to find $$\lim_{x\to 0}\frac{\tan^2 x+2x}{x+x^2}$$
This is what I did:
Let $f(x)=\frac{\tan^2(x)+2x}{x+x^2}=\frac{\frac{\sin^2(x)}{\cos^2(x)}+2x}{x+x^2}=\frac{\sin^2(x)+2x\cos^2(x)}{x(x+1)\cos^2(x)}=\frac{\sin^2(x)}{x(x+1)\cos^2(x)}+\frac{2}{x+1}=\frac{\sin (x)}{x}\cdot\frac{\sin(x)}{(x+1)\cos^2(x)}+\frac{2}{x+1}$
$\lim_{x\to0}\left(\frac{\sin(x)}{x}\cdot\frac{\sin(x)}{(x+1)\cos^2(x)}+\frac{2}{x+1}\right)=\underbrace{\left(\lim_{x\to0}\frac{\sin(x)}{x}\right)}_{=1}\cdot\left(\lim_{x\to0}\frac{\sin(x)}{(x+1)\cos^2(x)}\right)+\underbrace{\lim_{x\to0}\left(\frac{2}{x+1}\right)}_{=2}=\lim_{x\to0}\frac{\sin(x)}{(x+1)\cos^2(x)}+2=\frac{\lim_{x\to0}\sin(x)}{\underbrace{\left(\lim_{x\to0}x+1\right)}_{=1}\cdot\underbrace{\left(\lim_{x\to0}\cos^2(x)\right)}_{=1^2=1}}+2=\lim_{x\to0}\sin(x)+2=0+2=2$
Is this correct? I feel like I have assumed the some limits rather than proving it. We are allowed to use the common result $\frac{\sin x}{x}$ is $1$ as $x$ tends to $0$.
$$\frac{\tan^2 x+2x}{x+x^2}$$ is equivalent to$$\frac{\tan(x)\frac{\tan(x)}{x}+2}{1+x}$$