Useful concepts on limits of function - Basic mathematics(+2 level)

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 We usually denote a function by f(x).

What is a function ?

-> A function is nothing but a mapping or link of one location to another. If we consider these two locations as two non-empty sets X and Y, then a function f from X to Y is a rule of relation that assigns a unique element of Y to each element of X. The unique element of  Y corresponding to an element x of X (x`\in`X) assigned by f is denoted by f(x), so that we can write

y = f(x)

Symbol used in Mathematics, 

f: X `\rightarrow` Y

This symbol denotes that f is a function through which each element of Y is assigned to unique element of X. Elements of X, x's are called domain of y under f whereas elements y = f(x) are called image of x under f. 

Value of a function f:

Suppose f is a function from X to Y. Also,  x = a is an element in X (domain of f ) then the image f(a) which is an element of Y corresponding to x = a is said to be the value of function f at x = a. So, each value of f may be calculated by using corresponding domain. 

For example, if f(x) = `2x^3 + 3x + 5` is a function, then value of f at x = 1 is, 

`f(x=1) =  2 \times 1^3 + 3 \times 1 + 5`

f(x=1) = 10.

`\therefore` f(1) = 10, which is a value of f(x) at x = 1. 

In most cases, one has to determine value of function by considering it's limits.

Limits of Function:

  • A number M is said to be the limit of function f(x) at a point x = m if for every arbitrary chosen positive as well as very small number `\epsilon` , there exists corresponding number ẟ, greater than zero such that |f(x)-M| <  `\epsilon` for all values of x for which `m_o < |x-m|< ẟ`,  where |x-m| means the absolute value of x-m without any regard to sign. Symbolically, `\lim_(x -> m)` f(x) = M.
  • If x approaches to 'm' from right, that is, from larger values x than m, the limit of f(x) is called the right hand limit and is denoted as `\lim_( x->m^+ )` f(x).
  • If x approaches to m from left, that is, from the smaller values of x than m, the limit of f(x) is called the left hand limit denoted by `\lim_( x->m^- )`f(x). 

For example, if x has values as ..... -3 -2, -1, m, 1, 2, 3, ................ or

0.01, 0.02, 0.03, m, 0.06, 0.07, 0.08, then value of x approaching m from right side of m is called right hand limit, whereas those approaching from left side of m is are called left hand limit.

Some results of limit of functions:

If f and g are two real valued functions defined over certain domain, then

  • `\lim_( x->m )[f(x) \pm g(x)] = \lim_( x->m )f(x) \pm \lim_( x->m )g(x)`.
  • `\lim_(x->m)[f(x) \cdot g(x)] = \lim_(x->m)f(x) \cdot \lim_(x->m)g(x)`.
  • `\lim_(x->m) [\frac{f(x)}{g(x)}] = \frac{\lim_(x->m)f(x)}{\lim_(x->m)g(x)}` provided that `\lim_(x->m)g(x) \ne 0`.
  • `\lim_(x->m)[cf(x)] = c \times \lim_(x->m)f(x)` where c is a constant.
Meaning of infinity:
Usually,  infinity (`\infty`) means very large. But in all cases, this definition does not work. Value of `\infty` depends on condition under which is being used.
Consider a function, f(x) = `\frac{1}{x}`.
When x = 0, then f(x = 0) = `\frac{1}{0} = \infty`
In mathematics, division of any number by 0 has not been defined properly. So we can not exactly say how much large is `\infty`. 

When x = 3, f(x =3) = `\frac{1}{3}` = 0.33333333333 ...
 When x = 2, f(x = 2) = `\frac{1}{2}` = 0.5 
When x = 1, f(x = 1) = `\frac{1}{1} = 1`
When x = 0.9, f(x = 0.9) = `\frac{1}{0.9}` = 1.111111111111 ....
When x = 0.8, f(x = 0.8) = `\frac{1}{0.8} = 1.25`
Thus value of f(x) goes on increasing gradually, as value of x decreases. 
This means when value of x is minimum, then value of f(x) will be maximum. When value of x approaches to zero, then values of f(x) approaches to maximum (`\infty`). 
When x = 0.000000000001, f(x = 0.000000000001) = `\frac{1}{0.000000000001} = 10^12`, which is very large number.  However x is still greater than zero, so `10^12` is less than `\infty`.
This means `10^12` is very large value of f(x), but `10^12` is still less than `\infty`. Now you can think how big is the number `\infty`. 


L-Hospital's Rule: 

 L-Hospital's rule is a very famous and useful tool to solve problems of limit and continuity when there arises a case such as  `\frac{0}{0}`  or   `\frac{\infty}{\infty}`  to the limiting value of fraction of functions f(x) and g(x) given by  `\lim_(x->m)\frac{f(x)}{g(x)}`. These results are called indeterminate forms. 

So, for two functions f(x) and g(x),

If  `\lim_(x->m) \frac{f(x)}{g(x)} -> \frac{0}{0}` or `\frac{\infty}{\infty}`, then using L-Hospital's rule, we can have

`\lim_(x->m)\frac{f(x)}{g(x)} = \frac{f \prime (m)}{g \prime (m)}`.

Where f'(m) and g'(m) are functional values or limiting values of derivatives of f(x) and g(x) respectively.

Some Examples:

  • `\lim_(\theta ->0) \frac{sin \theta}{\theta} = 1`, `\theta` is angle meaured in radians.
  • `\lim_(\theta -> \infty) \frac{sin \theta}{\theta} = 0`.
  • `\lim_(x->0) \frac{e^x - 1}{x} =1`.
  • `\lim_(x->0) \frac{log(x +1) }{x} = 1` and so on.

Problems and their Solutions:

1.) Solve: `\lim_(x->0)\frac{e^x + e^(-x) - 2}{x^2}`. 

Solution:-

Here, 

L = `\lim_(x->0) \frac{e^x + e^(-x) - 2}{x^2}`

   = `\lim_(x->0)\frac{e^x + \frac{1}{e^x} - 2}{x^2}`

  =`\lim_(x->0) \frac{e^(2x) +1 - 2e^x}{x^2 e^x}` 

  = `\lim_(x->0)\frac{(e^x - 1)^2}{x^2 e^x}`

 = `\lim_(x->0) (\frac{e^x - 1}{x})^2 \cdot \lim_(x->0) \frac{1}{e^x}`

 = 1. `\frac{1}{e^0}`, `\therefore` `e^0 = 1`.

 = 1. 1

= 1. //

2.)  Solve: ` \lim_(\theta->0) \frac{cosec  \theta - cot \theta}{\theta}`.

Solution:-

Here,

L = `\lim_(\theta -> 0) \frac{cosec \theta - cot \theta}{\theta}`.

 = `\lim_(\theta -> 0) \frac{ \frac{1}{sin \theta} - \frac{cos \theta}{sin \theta}}{\theta}`.

=`\lim_(\theta -> 0) \frac{\frac{1- cos \theta}{sin \theta}}{\theta}`.

=`\lim_(\theta ->0) \frac{1 - cos \theta }{\theta \cdot sin \theta}`.

We need to express this to formula such that we can use `\lim_(\theta -> 0) \frac{sin \theta}{\theta} = 1`.

    `\therefore` L = `\lim_(\theta -> 0) \frac{1-cos \theta}{\theta^2 \cdot \frac{sin \theta}{\theta}}`.

L = `\lim_(\theta -> 0) \frac{1-cos \theta}{\theta^2} \cdot (\frac{1}{\lim_(\theta->0) \frac{sin \theta}{\theta}})`.

= `\lim_(\theta -> 0)\frac{1-cos \theta}{\theta^2} \cdot 1`

But `1 - cos \theta = 2sin^2 \frac{\theta}{2}`, So

L = `\lim_(\theta -> 0) \frac{2 sin^2 \frac{\theta}{2}}{\theta^2}`.

= `\lim_(\theta -> 0) \frac{2}{4} (\frac{sin \frac{\theta}{2}}{\frac{\theta}{2}})^2`.

= `\frac{1}{2} \cdot (\lim_(\theta->0)\frac{sin \frac{\theta}{2}}{\frac{\theta}{2}})^2`.

= `\frac{1}{2} \cdot 1`

= `\frac{1}{2}` //


To prove very important result in limits,

`\lim_(\theta -> 0) \frac{sin \theta}{\theta} =1`

Consider a circle MNX with center at O and radius r as shown in figure.

That is, OX = OM  = r.

XM is an arc that subtends angle `\theta` at  O. 

Let XY be a tangent at point X of circle which meets NM at Y. Then we join X to M and draw XZ perpendicular to NM. Then 

Area of  `\triangle`OXM   `\le` Area of sector AXM  `\le` Area of  `\triangle`OXY 

..........................(1)

Here,

Area of `\triangle` OXM = `\frac{1}{2} OM \cdot XZ` = `\frac{1}{2} r \cdot r sin\theta`.

                                         = `\frac{1}{2} r^2 sin\theta`.

Area of sector OXM = `\frac{1}{2} OX \cdot \overarc (XM)`

                                  = `\frac{1}{2} r \cdot r \theta`

                                 = `\frac{1}{2} r^2 \theta`.

Area of `\triangle` OXY = `\frac{1}{2} OX \cdot XY`

                                       = `\frac{1}{2} r \cdot r tan \theta`

                                      = `\frac{1}{2} r^2 tan \theta`.

Now, using these values in (1), we have

`\frac{1}{2} r^2 sin \theta \le  \frac{1}{2} r^2 \theta \le  \frac{1}{2} r^2 tan \theta`.

Here, `\frac{1}{2} r^2` cancels out from all terms. So,

` sin \theta \le \theta \le tan \theta`.

=> `sin \theta \le \theta \le \frac{sin \theta}{cos \theta}`

Dividing all sides by `sin \theta`, we get

`1 \le \frac{\theta}{sin \theta} \le \frac{1}{cos \theta}`

Taking reciprocal of all sides,

`1 \ge \frac{sin \theta}{\theta} \ge cos \theta`

Now taking limit as `\lim_(\theta -> 0)` on all sides,

`\lim_(\theta -> 0) 1 \ge \lim_(\theta -> 0) \frac{sin \theta}{\theta} \ge \lim_(\theta -> 0) cos \theta`

But `\lim_(\theta -> 0) cos \theta = 1`, So

`1 \ge \lim_(\theta -> 0) \frac{sin \theta}{\theta} \ge 1`

Since values of  `\lim_(\theta -> 0) \frac{sin \theta}{\theta}`  is greater and equal to 1 as well as smaller and equal to 1. So it must give us.

`\lim_(\theta ->0) \frac{sin \theta}{\theta} = 1`.

`\therefore \lim_(\theta ->0) \frac{sin \theta}{\theta} = 1`. Hence proved.

This formula can also be understood in this way,

For small `\theta`, `sin \theta \approx \theta`.

So for value of  `\theta` as smaller as 0 but not exactly 0, the ratio `\frac{sin \theta}{\theta}` becomes equal to 1. 

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