##### 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 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 1If $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$ 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 $\displaystyle P(x)={{a}_{n}}{{x}^{n}}+…….+{{a}_{1}}x+{{a}_{0}}$

$\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,P(x)=P(c)$

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 2If  $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,g(x)=m$ 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 3If $\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=l$ 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.