# Limits (Intuitive Introduction)

THE TANGENT PROBLEM
Given a function f(x) and a point $P(x_0, y_0)$ on its graph, find an equation of the line tangent to the graph at P

THE AREA PROBLEM
Given a function f, find the area between the graph of the f and an interval $[a,b]$ on the x-axis

A line is called tangent to a circle if it meets the circle at precisely one point.

But this definition is not satisfactory for other kind of curves, like

The line is tangent yet it meets the curve more than once.

The line occupying this limiting position we consider to be tangent line at P.
The areas of some plane regions can be calculated by subdividing them into finite number of rectangles or triangle, then adding the area of the constituent parts.

For many regions a more general approach is needed.

We approximate the area of this region by inscribing rectangles of equal width under the curve and adding the areas of these rectangles.

Our approximations will "approach" the exact area under the curve as a "limiting value".

Let us take
$f(x) = \frac{\sin(x)}{x}$
Where x is in radians.
Right hand limit
$\lim_{x \to 0^{+}}\frac{\sin x}{x}$
"The limits of $f(x)$ as $x$ approaches 0 from the right"

Left hand limit
$\lim_{x \to 0^{-}}\frac{\sin x}{x}$
"The limit of f(x) as x approaches 0 from left."
Using a calculator set to the radian mode, we have
 $x$ $f(x) = \frac{\sin x}{x}$ 1.0 0.84107 0.8 0.89670 0.6 0.94107 0.4 0.97355 0.2 0.99335 0.01 0.99998
 $x$ $f(x) = \frac{\sin x}{x}$ -1.0 0.84107 -0.8 0.89670 -0.6 0.94107 -0.4 0.97355 -0.2 0.99335 -0.01 0.99998

As x approach 0 from the left or right, f(x) approach 1.
The preceding ideas are summarized in this table.
 NOTATION HOW TO READ THE NOTATION $\lim_{x \to x_0^{+}} = L_1$ The limit of $f(x)$ as $x$ approaches $x_0$ from the right is equal to $L_1$ $\lim_{x \to x_0^{-}} = L_2$ The limit of f(x) as x approaches $x_0$ from the left is equal to $L_2$ $\lim_{x \to x_0} = L$ The limit of $f(x)$ as $x$ approaches $x_0$ is equal to $L$

Numerical Pitfalls
$f(x) = \sin(\frac{\phi}{x})$
$\lim_{x \to 0} \sin(\frac{\phi}{x}) = \lim_{x \to 0^+} \sin(\frac{\phi}{x}) = \lim_{x \to 0^{-}} \sin(\frac{\phi}{x})$
Table 2.4.4
 x f(x) x f(x) 1 0 -1 0 0.1 0 -0.1 0 0.01 0 -0.01 0 0.001 0 -0.001 0 0.0001 0 -0.0001 0

Existence of limits

$\lim_{x \to x_0^{-}} f(x) = \lim_{x \to x_0^{+}} f(x) = \lim_{x \to x_0} f(x) = +\infin$
$x$ approaches $x_0$ from left

$\lim_{x \to x_0^{-}} f(x) = \lim_{x \to x_0^{+}} f(x) = \lim_{x \to x_0} f(x) = -\infin$
$x$ approaches $x_0$ from right
Example

$\lim_{x \to x_0^{-}} f(x) = -\infin$       $\lim_{x \to x_0^{+}} f(x) = +\infin$
The limit of the function does not exist at $x_0$
Example

Here
$\lim_{x \to -\infin} f(x) = -1$       $\lim_{x \to +\infin} f(x) = 4$
So, limit of function does not exist.
Example

Here
as x → -∞, f(x) → +∞
as x → +∞, f(x) → -2
So, limit of function does not exist.
Example

Here
as x → -∞, f(x) → -∞
as x → +∞, f(x) → oscillates
So, limit of function does not exist.

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