\[ \lim_{x\to a} f(x)=L \Leftrightarrow \lim_{x\to a^-} f(x)=L=\lim_{x\to a^+} f(x) \]
\[ \lim_{x\to 0} |x| \]
\[ \lim_{x\to 0} \frac{|x|}{x} \]
\[ \lim_{x\to 0} \frac{x^2}{|x|} \]
\[ \lim_{x\to 0} \sqrt{x}=\]
\[ f(x)=\left\{ \begin{array}{ll} \sqrt{x-4} &, x>4\\ 8-2x &, x<4 \end{array} \right. \]
Assume
Then
\[ \lim_{x\to a} f(x)\leq \lim_{x\to a} g(x)\]
Strict inequality does not hold for the limits.
Ex. For \(x>0\),
\(\frac{1}{x}>0\)
But
\( \lim_{x\to \infty} \frac{1}{x}=0\)
Assume
Then
\[
\lim_{x\to a} g(x)=L
\]
\[ \lim_{x\to a} \left(x^2\cdot\sin(\frac{1}{x}) \right) \]
\[ \lim_{x\to a} x^2\cdot\sin(\frac{1}{x})=(\lim_{x\to a} x^2)\cdot\lim_{x\to a}(\sin(\frac{1}{x})) \] Why not?