Computing Limits - HMC Calculus Tutorial

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Computing Limits

Intuitively, we say that $\displaystyle \lim_{x\to c} f(x)=L$ if f is defined near (but not necessarily at) c and f(x) approaches L as x approaches c.

If we let x approach c from the left side only, we write $\displaystyle \lim_{x\to c^-} f(x)$ since x is approaching c from smaller values. Similarly, for x approaching c from the right, we write $\displaystyle \lim_{x\to c^+} f(x)$ . The two-sided limit $\displaystyle \lim_{x\to c} f(x)$ exists if and only if both of these one-sided limits exist and are equal.

An Intuitive Example

Consider the graph of a function f(x) shown below.

Evaluate each of the following. Select each one to check your reasoning.

		$\lim_{x \to -3} f(x)$
$\lim_{x \to 2^+} f(x)$	$\lim_{x \to -5^+} f(x)$	$\lim_{x \to -\infty} f(x)$
$\lim_{x \to 2^-} f(x)$	$\lim_{x \to -5^-} f(x)$	$\lim_{x \to \infty} f(x)$
$\lim_{x \to 2} f(x)$	$\lim_{x \to -5} f(x)$	$\lim_{x \to c} f(x) \textrm{~for~} c\neq -5, -3, 2.$

Definition of the Limit

More rigorously, let f be defined at all x in an open interval containing c, except possibly at c itself.

Then

$\begin{displaymath}\lim_{x\to c} f(x)=L\end{displaymath}$

if and only if for each ε > 0, there exists a δ > 0 such that

$\begin{displaymath}\textrm{if~} 0<\vert x-c\vert<\delta \textrm{~then~}\vert f(x)-L\vert<\varepsilon.\end{displaymath}$

Figures

Exploration
In words, $\displaystyle \lim_{x\to c} f(x)=L$ if and only if by taking x close enough to c we can get f(x) arbitrarily close to L.

Properties of the Limit

Each of the following properties is proven using the rigorous definition of the limit. Let $\lim$ stand for $\displaystyle \lim_{x\to c}$ , $\displaystyle \lim_{x\to c^+}$ , or $\displaystyle \lim_{x\to c^-}$ . Assume $\lim f(x)$ and $\lim g(x)$ both exist.

(Uniqueness) If $\lim f(x)=L_1$ and $\lim f(x)=L_2$ , then .
(Addition) $\lim [f(x)+g(x)]=\lim f(x) +\lim g(x)$ .
(Scalar multiplication) $\lim [\alpha f(x)]=\alpha \lim f(x)$ .
(Multiplication) $\lim [f(x)g(x)]=\lim f(x) \cdot \lim g(x)$ .
(Division) $\displaystyle \lim \frac{f(x)}{g(x)}=\frac{\lim f(x)}{\lim g(x)}$ , provided $\lim g(x)\neq 0$ .
(Powers) $\displaystyle \lim [f(x)]^n=[\lim f(x)]^n$ for any positive integer .

In practice, much of the time we can ``reason out'' the value of a limit without explicitly using the ε-δ definition.

Examples

$\displaystyle\lim_{x\to 2} \sqrt{x^2+12}=4$ since the function $f(x)=\sqrt{x^2+12}$ is continuous at x=2 and f(2)=4.
$\displaystyle\lim_{x\to \infty} \frac{1}{x}=0$ since as x increases, $\displaystyle \frac{1}{x}$ gets arbitrarily close to 0.
$\displaystyle\lim_{x\to 0^+} \ln \vert x\vert$ tends to -∞ and so does not exist since as x decreases to 0, ln |x| gets arbitrarily large in magnitude and negative.
$\displaystyle\lim_{x\to 3} \frac{x^2-9}{x-3}=6$ even though $\displaystyle f(x)=\frac{x^2-9}{x-3}$ is undefined at x=3 since $\displaystyle \frac{x^2-9}{x-3}=x+3$ and $\displaystyle \lim_{x\to 3} x+3 =6$ .

What about something like $\displaystyle\lim_{x\to 0} \frac{\sin x}{x}$ ? When we cannot easily ``reason out'' the value of a limit, we can often use numerical methods or L'Hôpital's Rule to determine the value of the limit. Can you convince yourself that $\displaystyle\lim_{x\to 0} \frac{\sin x}{x}=1$ ?

Key Concept

Let the function f be defined at all x in an open interval containing c, except possibly at c itself.

Then

$\begin{displaymath}\lim_{x\to c} f(x)=L\end{displaymath}$

if and only if for each ε > 0, there exists a δ > 0 such that

$\begin{displaymath}\textrm{if~} 0<\vert x-c\vert<\delta \textrm{~then~}\vert f(x)-L\vert<\varepsilon.\end{displaymath}$

In words, the limit of f(x) as x approaches c is L if and only if by taking x close enough to c we can get f(x) arbitrarily close to L.

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