If two functions go to infinity at zero, does the difference go to zero? If $\lim_{ x\to0} f(x) = \infty$ and $\lim_{ x\to0} g(x) = \infty$, then $\lim_{ x\to0} [f(x) − g(x)] = 0$. 
 True or False??
 A: No
Let $f(x)=\dfrac{3m}{\sin^2x}$ and $g(x)=\dfrac{3m}{x^2}$
$$\lim_{x\to0}[f(x)-g(x)]\neq0 \quad,\forall m\in \mathbb N\cap\{m\ge 1\}$$
Edit:$$\lim_{x\to0}[f(x)-g(x)]=\lim_{x\to0}\left[\dfrac{3m}{\sin^2x} -\dfrac{3m}{x^2}\right]=\lim_{x\to0}\left[\dfrac{mx^4}{x^4}\right]$$
A: \begin{align}
{\rm f}\left(x\right) = {1 \over x}\,,
\quad
{\rm g}\left(x\right) = {1 \over x}\,;
& \qquad\qquad{\large\mbox{( TRUE )}}
\\[3mm]
{\rm f}\left(x\right) = {2 \over x}\,,
\quad
{\rm g}\left(x\right) = {1 \over x}\,;
& \qquad\qquad{\large\mbox{( FALSE )}}
\end{align}
A: False. Let $f(x)=\dfrac1{x^2}$ and $g(x)=\dfrac1{x^2} + 1$.
A: False! 
For example 
$$
f(x)=\frac2{x}, \ g(x)=\frac1{x}.
$$
A: Blurry, as the examples have shown you, the essense here is that you cannot work with infinity as if it is a number. Infinity is an entity with which you cannot calculate in a normal fashion as with ordinary numbers. Just like saying infinity divided by infinity is equal to 1. Very often that is not true either. In case of infinity - infinity, usually making one term (most of the time fractional) out of them is the way to go.
Feel free to post an actual limit problem and we will help you. 
