# What are some sample limit problems?

Aug 17, 2017

A few examples...

#### Explanation:

Limit problems can take many forms, so I will just give a few examples in increasing order of complexity...

$\textcolor{w h i t e}{}$
Polynomials

Polynomials are always defined and continuous everywhere. Hence if $f \left(x\right)$ is a polynomial and $a \in \mathbb{R}$, then:

${\lim}_{x \to a} f \left(x\right) = f \left(a\right)$

The behaviour as $x \to + \infty$ or $x \to - \infty$ is determined solely by the leading term (i.e. term of highest degree), which will dominate the other terms when $x$ is sufficiently large.

Hence we get rules:

• If $f \left(x\right)$ is of odd degree with positive leading coefficient then:$\textcolor{w h i t e}{\frac{0}{0}}$
${\lim}_{x \to - \infty} f \left(x\right) = - \infty$ and ${\lim}_{x \to + \infty} f \left(x\right) = + \infty$

• If $f \left(x\right)$ is of odd degree with negative leading coefficient then:$\textcolor{w h i t e}{\frac{0}{0}}$
${\lim}_{x \to - \infty} f \left(x\right) = + \infty$ and ${\lim}_{x \to - \infty} f \left(x\right) = - \infty$

• If $f \left(x\right)$ is of even degree with positive leading coefficient then:$\textcolor{w h i t e}{\frac{0}{0}}$
${\lim}_{x \to - \infty} f \left(x\right) = + \infty$ and ${\lim}_{x \to + \infty} f \left(x\right) = + \infty$

• If $f \left(x\right)$ is of even degree with negative leading coefficient then:$\textcolor{w h i t e}{\frac{0}{0}}$
${\lim}_{x \to - \infty} f \left(x\right) = - \infty$ and ${\lim}_{x \to + \infty} f \left(x\right) = - \infty$

Example:

Given $f \left(x\right) = - 2 {x}^{3} + 5 x + 7$ evaluate:

• ${\lim}_{x \to 1} f \left(x\right)$

• ${\lim}_{x \to - \infty} f \left(x\right)$

• ${\lim}_{x \to + \infty} f \left(x\right)$

$\textcolor{w h i t e}{}$
Rational functions

Like polynomials these are continuous everywhere, except when the denominator is zero. Typical problems might involve evaluating a limit where both numerator and denominator tend to $0$. These usually involve simplifying the rational expression by identifying common factors.

Example:

Given:

$f \left(x\right) = \frac{{x}^{2} - 1}{{x}^{2} + x - 2}$

what is ${\lim}_{x \to 1} f \left(x\right)$ ?

We find:

$f \left(x\right) = \frac{{x}^{2} - 1}{{x}^{2} + x - 2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x + 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x + 2\right)} = \frac{x + 1}{x + 2}$

with exclusion $x \ne 1$

So:

${\lim}_{x \to 1} f \left(x\right) = {\lim}_{x \to 1} \frac{x + 1}{x + 2} = \frac{\textcolor{b l u e}{1} + 1}{\textcolor{b l u e}{1} + 2} = \frac{2}{3}$

$\textcolor{w h i t e}{}$

Example:

Given:

$f \left(x\right) = \left(\sqrt{x + 1} - \sqrt{x - 1}\right) \sqrt{x}$

What is ${\lim}_{x \to \infty} f \left(x\right)$ ?

We find:

${\lim}_{x \to \infty} \left(\sqrt{x + 1} - \sqrt{x - 1}\right) \sqrt{x}$

$= {\lim}_{x \to \infty} \frac{\left(\sqrt{x + 1} - \sqrt{x - 1}\right) \left(\sqrt{x + 1} + \sqrt{x - 1}\right) \sqrt{x}}{\sqrt{x + 1} + \sqrt{x - 1}}$

$= {\lim}_{x \to \infty} \frac{\left(\left(x + 1\right) - \left(x - 1\right)\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{x}}}}}{\left(\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{x}}}}}$

$= {\lim}_{x \to \infty} \frac{2}{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}}$

$= \frac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}}$

$= \frac{2}{2}$

$= 1$