Limit

An example of limit

If a variable $x$ takes values which are more and more close to a finite number $a$ , then we say that $x$ approaches $a$ written as $x\to a$).

• If values of $x$ come closer to $a$ but are always greater than $a$ , then we say that $x$ approaches $a$ form right ($x\to a^+$).
• If values of $x$ come closer to $a$ but are always less than $a$ , then $x$ approaches $a$ from left ($x\to a^-$) .

The concept of a limit is essentially what separates the field of calculus, and analysis in general, from other fields of mathematics such as geometry or algebra.

The concept of a limit may apply to:

• Limit of a function
• Limit of a sequence
• Limit of a series

Examples

If $x\to2$ , then $x$ can approach to '2' from two sides:

• From right side: In notation we write $x\to2^+$ means $x$ is coming closer to '2' from right i.e. it is more than '2'.

$\begin{align}&x=2.1\\&x=2.01\\&x=2.0001\\&\vdots\end{align}$

• From left side: In notation we write $x\to2^-$ mean $x$ is coming closer to 2 from left i.e. it is less than '2'.

$\begin{align}&x=1.9\\&x=1.99\\&x=1.9999\\&\vdots\end{align}$

Meaning of a limiting value

Let $f(x)$ be function of $x$ . If the expression $f(x)$ comes close and stays close to $L$ as $x$ approaches $a$ then we say that $L$ is the limit of $f(x)$ as $x$ approaches $a$ . The value $x$converges to is called the limit point.

In notation, it is written as $\lim_{x\to a}f(x)=L$ .

Right hand limit

If $f(x)$ approaches $L_1$ as $x$ approaches $a$ from the right, then $L_1$ is called as the right hand limit of $f(x)$ .

Right hand limit can be expressed in two ways:-

• $\lim_{x\to a^+}f(x)=L_1$
• $\lim_{h\to0}f(a+h)\text{ Put }x=a+h\text{ in the above result}$

Left hand limit

If $f(x)$ approaches $a$from the left, then $L_2$ is called the left hand limit of $f(x)$ . The left hand limit can be expressed in two ways:-

• $\lim_{x\to a^-}f(x)=L_2$
• $\lim_{h\to0}f(a-h)\text{ Put }x=a-h\text{ in the above result}$

Note that $h$ is a positive real number, that we let approach 0..

Existence of limit

For existence of limit at $x=a$

$\begin{align}&\Rightarrow LHL=RHL\\ &\Rightarrow L_1=L_2\\ &\Rightarrow\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)\end{align}$

Illustrating the concept

If $f(x)=\frac{x^2-4}{x-2}$ , then evaluate $\lim_{x\to2}f(x)$ .

L.H.L. = $\lim_{x\to2^-}\frac{x^2-4}{x-2}$ i.e. $x$ is coming closer to 2 but it is less than '2'. So, observe the situation in table below:

$x$	$2-x$	$f(x)$
1.9	0.1	3.9
1.99	0.01	3.99
1.999	0.001	3.999
$\vdots$	$\vdots$	$\vdots$
Coming closer to 2 but less than 2		Coming closer to 4 but less than 4

Infinity in limits

Limits can approach infinity:

$\lim_{x \to \infty} \frac {2x}{x-1} = 2$

If a limit approaching infinity is a real value, then as the variable gets larger and larger, the limit converges.

Limits can also be equal infinity:

$\lim_{x\to0}-\csc^2(x) = -\infty$

On a graph, an infinite limit would appear to skyrocket or plummet to the top and bottom edges of the graph (asymptote).

Calculating limits

Direct substitution

If the function is continuous, limits can be evaluated by directly substituting the variable with the value it approaches. For example, because the cubic is continuous we have

$\lim_{x \to 6}{x^3} = (6)^3 = 216$

In the limit below, direct substitution will give a false answer.

$\begin{align} \lim_{x \to \pi} e^{cot(x)} & \neq e^{cot(\pi)} \\ \lim_{x \to \pi^-} e^{cot(x)} & = 0 \\ \lim_{x \to \pi^+} e^{cot(x)} & = \infty \end{align}$

Graphing will show that $e^{cot(x)}$is not continuous at $x = \pi$, and that therefore $\lim_{x \to \pi} e^{cot(x)}$does not exist.

Indeterminate form

When a discontinuous function is improperly computed using direct substitution, one of the following peculiar results may be concluded:

$\frac 00 \qquad \frac \infty\infty \qquad 0*\infty \qquad 0^0 \qquad \infty^0 \qquad 1^\infty \qquad \infty-\infty$

The indeterminate forms do not provide enough information to identify the real limit.

The limits below are unequal. If they are directly substituted, they reach an indeterminate form and cannot be distinguished.

$\begin{align} \lim_{x \to 0} \frac xx & = 1 \\ \lim_{x \to 0} \frac {x^2}x & = 0 \\ \lim_{x \to 0} \frac xx & \neq \lim_{x \to 0} \frac {x^2}x \\ \frac {(0)}{(0)} & \neq \frac {(0)^2}{(0)} \\ \frac 00 & \neq \frac 00 \end{align}$

There are some techniques to deal with limits with indeterminate form:

Algebraic manipulation

Discontinuities can be eliminated through algebraic manipulation:

$\lim _{x\to -1}{\frac {x^{2}-1}{x+1}}=\lim _{x\to -1}{\frac {(x+1)(x-1)}{x+1}}=\lim _{x\to -1}x-1=-2$

Knowing the well-known limits

Here're some well-known limits:

1. $\lim_{x\to0}\frac{\sin(x)}{x}=1$
2. $\lim_{x \to 0}\frac{e^x - 1}{x} = 1$
3. $\lim_{x \to \infty}(1 + \frac{1}{x})^x = e$

L'Hôpital's rule

L'Hôpital's rule states that if $f(x)=g(x)=0$, $\frac{d}{dx}f(x)$ and $\frac{d}{dx}g(x)$ exist then:

$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$

For example, to compute $\lim _{x\to 0}{\frac {e^{x}-\cos(x)}{x}}$, first plug 0 to the numerator and denominator of the limit:

• numerator $=e^{0}-\cos(0)=1-1=0$
• denominator $= 0$

When x is 0, both the numerator and denominator are 0, meaning that the L'Hôpital's rule can be used on both of them:

$\lim _{x\to 0}{\frac {e^{x}-\cos(x)}{x}}=\lim _{x\to 0}{\frac {{\frac {d}{dx}}(e^{x}-\cos(x))}{{\frac {d}{dx}}(x)}}=\lim _{x\to 0}e^{x}+\sin(x)=e^{0}+\sin(0)=1$