Suppose \(c\) is a constant and the limits
\[\lim_{x\to a} f(x)~and~\lim_{x\to a} g(x)\]
exist. Then
\(\displaystyle\lim_{x\to a} [f(x)+g(x)]=\lim_{x\to a} f(x)+\lim_{x\to a} g(x)\)
\(\displaystyle\lim_{x\to a} [f(x)-g(x)]=\lim_{x\to a} f(x)-\lim_{x\to a} g(x)\)
\(\displaystyle\lim_{x\to a} [ c\cdot g(x)]=c \cdot \lim_{x\to a} g(x)\)
\(\displaystyle\lim_{x\to a} [f(x) \cdot g(x)]=\lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x)\)
\(\displaystyle\lim_{x\to a} \frac{f(x)}{g(x)}=\frac{\ds \lim_{x\to a} f(x)}{\ds \lim_{x\to a} g(x)}~if~\lim_{x\to a} g(x)\neq0\)
Do not forget \(\displaystyle\lim_{x\to a} g(x)\neq0\) for law 5.
\[\displaystyle\lim_{x\to a} \frac{f(x)}{g(x)}=\frac{\ds \lim_{x\to a} f(x)}{\ds \lim_{x\to a} g(x)}~if~\lim_{x\to a} g(x)\neq0\]
One-sided limits also have similar rules.
Using the Limit Laws and the graphs of \(f\) and \(g\) to evaluate the following limits, if they exists. If it does not exists, check the infinite limits.
Since \(\lim_{x\to 2} f(x)\approx 1.5\neq0\) and \( \lim_{x\to 2} g(x)=0\),
it may give an infinite limit. So we need to check one-sided limits.
If \( x ~< 2 \) and \( x \) is very close to \(2\), \(f(x)~>0,~g(x)<0\).
\[ \lim_{x\to 2^-} \frac{f(x)}{g(x)}=-\infty \]
If \(x>2\) and \(x\) is very close to \( 2\), \(f(x)>0,~g(x)>0\).
\[ \lim_{x\to 2^+} \frac{f(x)}{g(x)}=+\infty \]
Let \(\displaystyle\lim_{x\to 1^-} f(x)=2,~\lim_{x\to 1^+} f(x)=-1\) \(,~\lim_{x\to 1^-} g(x)=-3, ~\lim_{x\to 1^+} g(x)=0\).
Find
\[\displaystyle\lim_{x\to 1} [f(x)+g(x)]\]
Let \(f\) is a polynomial or a rational function or radical function.
If a is in the interior of domain of \(f\),
\[ \lim_ {x\to a} f(x)=f(a) \]
If \(a\) is in the boundary of the domain,
\[ \lim_ {x\to a^{\pm}} f(x)=f(a) \] depending on the side of boundary.
(Namely, \(f(x)\) is continuous on the domain.)
Ex. \(f(x)=\sqrt{x}\) and \(Domain(f)=[0,\infty)\)
\[ \lim_ {x\to a}f(x)=f(0)=0,~for~a>0\]
\[ \lim_ {x\to 0^{+}}f(x)=f(0)=0\]
\[ \lim_ {x\to 5} (2x^2-3x+4) \]
\[ \lim_ {x\to -2} \frac{x^3+2x^2-1}{5-3x} \]
If \( f(x)=g(x) \) when \( x\neq a\), then
\[ \lim_ {x \to a} f(x)=\lim_{x\to a} g(x) \]
if the limits exist.
\[ \lim_ {h\to 0} \frac{(3+h)^2-9}{h} \]
\[ \lim_ {t\to 0} \frac{\sqrt{t^2+9}-3}{t^2} \]