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. Click each one to check your reasoning.
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 \[\lim_{x\to c} f(x)=L\] if and only if for each $\varepsilon >0$, there exists a $\delta >0$ such that \[{\small\textrm{if }} 0<|x-c|<\delta {\small\textrm{ then }}|f(x)-L|<\varepsilon.\] 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 $L_1=L_2$.
- 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 $n$.
In practice, much of the time we can “reason out” the value of a limit without explicitly using the $\varepsilon$–$\delta$ definition.
Example
- $\displaystyle\lim_{x\to 2} \sqrt{x^2+12}=4$ 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 $0$.
- $\displaystyle\lim_{x\to 0^+} \ln |x|$ tends to $-\infty$ 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 $3$ 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 Concepts
Let $f$ be defined at all $x$ in an open interval containing $c$, except possibly at $c$ itself.
Then
\[\lim_{x\to c} f(x)=L\]
if and only if for each $\varepsilon >0$, there exists a $\delta >0$
such that
\[{\small\textrm{if }} 0<|x-c|<\delta {\small\textrm{ then }}|f(x)-L|<\varepsilon.\]
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 $L$.