Theorems on Limits
It can be rather tedious to apply the $\displaystyle \varepsilon $ and $\displaystyle \delta $ limit test to individual functions. By remembering some basic theorems about limits we can avoid the some of this repetitive work.
We shouldn’t forget that if a limit exists it is always unique.
“The Uniqueness of a Limit”
If $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$ and $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=m$, then l=m
Theorem 1: If $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$ and $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=m$, then:
(i) $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=\left[ {f(x)+g(x)} \right]=l+m$
(ii) $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=\left[ {af(x)} \right]=al$ for each real $\displaystyle a$
(iii) $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\left[ {f(x)g(x)} \right]=lm$
The results of theorem 1 are easily extended to any finite collection of functions, namely if $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,{{f}_{1}}(x)={{l}_{1}}$, $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,{{f}_{2}}(x)={{l}_{2}}$ ……………$\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,{{f}_{n}}(x)={{l}_{n}}$ then:
1.1 $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\left[ {{{a}_{1}}{{f}_{1}}(x)+{{a}_{2}}{{f}_{2}}(x)+…….+{{a}_{n}}{{f}_{n}}(x)} \right]={{a}_{1}}{{l}_{1}}+{{a}_{2}}{{l}_{2}}+…….+{{a}_{n}}{{l}_{n}}$
1.2 $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\left[ {{{f}_{1}}(x){{f}_{2}}(x)…….{{f}_{n}}(x)} \right]={{l}_{1}}{{l}_{2}}……..{{l}_{n}}$
Don’t Forget!
For every polynomial $\displaystyle P(x)={{a}_{n}}{{x}^{n}}+…….+{{a}_{1}}x+{{a}_{0}}$
$\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,P(x)=P(c)$
TIP! See also Worked Examples – Limits
Example 1: Find the limits:
a) $ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,(5{{x}^{2}}-12x+2)$
b) $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,(14{{x}^{5}}-7{{x}^{2}}+2x+8)$
c) $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,(2{{x}^{3}}+{{x}^{2}}-2x-3)$
Solution: Applying theorem 1 and the rule for polynomials we get:
a) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,(5{{x}^{2}}-12x+2)=5\cdot {{(1)}^{2}}-12\cdot (1)+2=-5$
b) $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,(14{{x}^{5}}-7{{x}^{2}}+2x+8)=14\cdot {{(0)}^{5}}-7\cdot {{(0)}^{2}}+2\cdot (0)+8=8$
c) $ \displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,(2{{x}^{3}}+{{x}^{2}}-2x-3)=2\cdot {{(-1)}^{3}}+{{(-1)}^{2}}-2\cdot (-1)-3=-2$
Theorem 2: If $ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,g(x)=m$ with $\displaystyle m\ne 0$, then $ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\frac{1}{{g(x)}}=\frac{1}{m}$
Example 2: Find the limits:
a) $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{1}{{{{x}^{2}}}}$
b) $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{1}{{{{x}^{3}}-1}}$
c) $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,\frac{1}{{\left| x \right|}}$
Soulution: Applying theorem 2:
a) $ \displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{1}{{{{x}^{2}}}}=\frac{1}{{{{{(4)}}^{2}}}}=\frac{1}{{16}}$
b) $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{1}{{{{x}^{3}}-1}}=$
$\displaystyle \frac{1}{{{{{(2)}}^{3}}-1}}=\frac{1}{{8-1}}=\frac{1}{7}$
c) $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,\frac{1}{{\left| x \right|}}=$
$\displaystyle \frac{1}{{\left| {-3} \right|}}=\frac{1}{3}$
Theorem 3: If $ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$ and $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,g(x)=m$ with $\displaystyle m\ne 0$, then $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\frac{{f(x)}}{{g(x)}}=\frac{l}{m}$
Don’t forget!
If P and Q are polynomials and $\displaystyle Q(x)\ne 0$ then $ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\frac{{P(x)}}{{Q(x)}}=\frac{{P(c)}}{{Q(c)}}$
Example 3: Find the limits:
a) $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{{3x-5}}{{{{x}^{2}}+1}}$
b) $\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-3{{x}^{2}}}}{{1-{{x}^{2}}}}$
Solutions: Applying theorem 3 and the rule for polynomials on quotients we get:
a) $ \displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{{3x-5}}{{{{x}^{2}}+1}}=\frac{{3\cdot 2-5}}{{{{{(2)}}^{2}}+1}}=\frac{1}{5}$
b) $\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-3{{x}^{2}}}}{{1-{{x}^{2}}}}=\frac{{{{{(3)}}^{3}}-3\cdot {{{(3)}}^{2}}}}{{1-{{{(3)}}^{2}}}}=\frac{{27-27}}{{1-9}}=0$
There is no point looking for a limit that does not exist. The next theorem gives a condition under which a quotient does not have a limit.
Theorem 4: If $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$ with $\displaystyle l\ne 0$ and $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,g(x)=0$, then $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\frac{{f(x)}}{{g(x)}}$ does not exist.
Example 4: Evaluate the limits that exist:
a) $ \displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-x-6}}{{x-3}}$
b) $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{{{{({{x}^{2}}-3x-4)}}^{2}}}}{{x-4}}$
c) $ \displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,\frac{{x+1}}{{{{{(2{{x}^{2}}+7x+5)}}^{2}}}}$
Solution: In each case both numerator and denominator tend to zero, and so we have to be careful.
a) First we factor the numerator:
$ \displaystyle \frac{{{{x}^{2}}-3x-6}}{{x-3}}=\frac{{(x+2)(x-3)}}{{x-3}}$
For $\displaystyle x\ne 3$, $ \displaystyle \frac{{{{x}^{2}}-3x-6}}{{x-3}}=x+2$
Thus, $ \displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-x-6}}{{x-3}}=\underset{{x\to 3}}{\mathop{{\lim }}}\,(x+2)=5$
b) Note that $\displaystyle \frac{{{{{({{x}^{2}}-3x-4)}}^{2}}}}{{x-4}}=$
$\displaystyle \frac{{{{{\left[ {(x+1)(x-4)} \right]}}^{2}}}}{{x-4}}=$$\displaystyle \frac{{{{{(x+1)}}^{2}}{{{(x-4)}}^{2}}}}{{x-4}}$
So that, for $\displaystyle x\ne 4$, $\displaystyle \frac{{{{{({{x}^{2}}-3x-4)}}^{2}}}}{{x-4}}={{(x+1)}^{2}}(x-4)$
It follows then that $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{({{x}^{2}}-3x-4)}}{{x-4}}=\underset{{x\to 4}}{\mathop{{\lim }}}\,{{(x+1)}^{2}}(x-4)=0$
c) Since $ \displaystyle \frac{{x+1}}{{{{{(2{{x}^{2}}+7x+5)}}^{2}}}}=$
$\displaystyle \frac{{x+1}}{{{{{\left[ {(2x+5)(x+1)} \right]}}^{2}}}}=$$ \displaystyle \frac{{x+1}}{{{{{(2x+5)}}^{2}}{{{(x+1)}}^{2}}}}$
You can see that, for $\displaystyle x\ne -1$
$\displaystyle \frac{{x+1}}{{{{{(2{{x}^{2}}+7x+5)}}^{2}}}}=\frac{1}{{{{{(2x+5)}}^{2}}(x+1)}}$
By theorem 4 $ \displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,\frac{1}{{{{{(2x+5)}}^{2}}(x+1)}}$ does not exist.
Because the limit of the denominator tends again to zero and threfore,
$\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,\frac{{x+1}}{{{{{(2{{x}^{2}}+7x+5)}}^{2}}}}$ does not exist either.
