How to prove that $\lim\limits_{x\to 0}\frac{\tan^2{x}}{x^2+x}=0$ Basically, I need to calculate this
$$\lim_{x\to 0} \frac{\tan^2{x}}{x^2+x}$$
However, I'm not supposed to use L'Hopital's rule. I feel like squeeze theorem could be helpful but I can't find an adequate trigonometric property just yet. Any suggestions? 
Thanks in advance. 
 A: Rewrite the function as $\left(\frac{\tan x}{x}\right)^2\frac{x}{x+1}$.
A: You have
$$\begin{equation}\begin{aligned}
\lim_{x\to 0} \frac{\tan^2{x}}{x^2+x} & = \lim_{x\to 0} \frac{\sin^2{x}}{\cos^2{x}(x^2)(1 + \frac{1}{x})} \\
& = \lim_{x\to 0} \left(\frac{\sin{x}}{x}\right)^2\left(\frac{1}{\cos^2{x}}\right)\left(\frac{1}{1 + \frac{1}{x}}\right) \\
& = 1(1)(0) \\
& = 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note this uses the fairly well known $\lim\limits_{x\to 0} \left(\frac{\sin{x}}{x}\right) = 1$ limit (e.g., as shown in the Trigonometric functions section of Wikipedia's "List of limits" article).
A: $$\lim_{x\to 0} \frac{\tan^2{x}}{x^2+x}=\lim_{x\to 0} \frac{\sin^2{x}}{\cos ^2 x(x^2+x)}$$
Note that $cos x \to 1$ as $x\to 0$ therefore, $$\lim_{x\to 0} \frac{\sin^2{x}}{\cos ^2 x(x^2+x)}= \lim_{x\to 0} \frac{\sin^2{x}}{(x^2+x)}=$$
$$\lim_{x\to 0} \frac{\sin^2{x}}{x^2}.\frac {x}{(x+1)}=0$$
A: For small positive $x$,
$$ \tan^2(x) = \frac{ \sin^2 (x) }{ \cos^2 (x)} \leq \frac{x^2}{\cos^2 (x)} \leq 2 \cdot x^2,$$
$$ \frac{1}{x^2 + x} \leq \frac{1}{x},$$
and furthermore $\frac{ \tan^2 (x) }{ x^2 + x } \geq 0$.
Proceed by squeeze theorem, as you stated.
A: We have that
$$\frac{\tan^2{x}}{x^2+x}=\frac{\tan^2{x}}{x^2}\frac{x^2}{x^2+x}=\frac{\tan^2{x}}{x^2}\frac{x}{x+1} \to 1 \cdot 0=0$$
A: Hint: Use the fact that ${\displaystyle \lim_{x\rightarrow0}\frac{\sin x}{x}}=1$ to verify that ${\displaystyle \lim_{x\rightarrow0}\frac{\tan x}{x}}=1$ as well. Then rewrite $\dfrac{\tan^{2}x}{x^{2}+x}=\dfrac{\tan x}{x}\cdot\dfrac{\tan x}{x+1}$ and take the limit.
A: In a product, you can replace $\tan x$ by $x$, and
$$\frac{x^2}{x^2+x}=\frac x{x+1}$$ tends to $0$.
