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

 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∈X) assigned by f is denoted by f(x), so that we can write

y = f(x)

Symbol used in Mathematics, 

f: X → 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) = 2x3+3x+5 is a function, then value of f at x = 1 is, 

f(x=1)= 2×13+3×1+5

f(x=1) = 10.

∴ 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 ε , there exists corresponding number ẟ, greater than zero such that |f(x)-M| <  ε for all values of x for which mo<|x-m|<ẟ,  where |x-m| means the absolute value of x-m without any regard to sign. Symbolically, limx→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 limx→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 limx→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

  • limx→m[f(x)±g(x)]=limx→mf(x)±limx→mg(x).
  • limx→m[f(x)â‹…g(x)]=limx→mf(x)â‹…limx→mg(x).
  • limx→m[f(x)g(x)]=limx→mf(x)limx→mg(x) provided that limx→mg(x)≠0.
  • limx→m[cf(x)]=c×limx→mf(x) where c is a constant.
Meaning of infinity:
Usually,  infinity (∞) means very large. But in all cases, this definition does not work. Value of ∞ depends on condition under which is being used.
Consider a function, f(x) = 1x.
When x = 0, then f(x = 0) = 10=∞
In mathematics, division of any number by 0 has not been defined properly. So we can not exactly say how much large is ∞

When x = 3, f(x =3) = 13 = 0.33333333333 ...
 When x = 2, f(x = 2) = 12 = 0.5 
When x = 1, f(x = 1) = 11=1
When x = 0.9, f(x = 0.9) = 10.9 = 1.111111111111 ....
When x = 0.8, f(x = 0.8) = 10.8=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 (∞). 
When x = 0.000000000001, f(x = 0.000000000001) = 10.000000000001=1012, which is very large number.  However x is still greater than zero, so 1012 is less than ∞.
This means 1012 is very large value of f(x), but 1012 is still less than ∞. Now you can think how big is the number ∞


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  00  or   ∞∞  to the limiting value of fraction of functions f(x) and g(x) given by  limx→mf(x)g(x). These results are called indeterminate forms. 

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

If  limx→mf(x)g(x)→00 or ∞∞, then using L-Hospital's rule, we can have

limx→mf(x)g(x)=f′(m)g′(m).

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

Some Examples:

  • limθ→0sinθθ=1, θ is angle meaured in radians.
  • limθ→∞sinθθ=0.
  • limx→0ex-1x=1.
  • limx→0log(x+1)x=1 and so on.

Problems and their Solutions:

1.) Solve: limx→0ex+e-x-2x2

Solution:-

Here, 

L = limx→0ex+e-x-2x2

   = limx→0ex+1ex-2x2

  =limx→0e2x+1-2exx2ex 

  = limx→0(ex-1)2x2ex

 = limx→0(ex-1x)2â‹…limx→01ex

 = 1. 1e0, ∴ e0=1.

 = 1. 1

= 1. //

2.)  Solve: limθ→0cosec Î¸-cotθθ.

Solution:-

Here,

L = limθ→0cosecθ-cotθθ.

 = limθ→01sinθ-cosθsinθθ.

=limθ→01-cosθsinθθ.

=limθ→01-cosθθ⋅sinθ.

We need to express this to formula such that we can use limθ→0sinθθ=1.

    ∴ L = limθ→01-cosθθ2â‹…sinθθ.

L = limθ→01-cosθθ2⋅(1limθ→0sinθθ).

= limθ→01-cosθθ2⋅1

But 1-cosθ=2sin2θ2, So

L = limθ→02sin2θ2θ2.

= limθ→024(sinθ2θ2)2.

= 12⋅(limθ→0sinθ2θ2)2.

= 12â‹…1

= 12 //


To prove very important result in limits,

limθ→0sinθθ=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 θ 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  â–³OXM   ≤ Area of sector AXM  ≤ Area of  â–³OXY 

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

Here,

Area of △ OXM = 12OM⋅XZ = 12r⋅rsinθ.

                                         = 12r2sinθ.

Area of sector OXM = 12OXâ‹…

                                  = \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. 

----------------------

Thanks for the visit !

😊😊😊😊😊

Do Follow for more updates

----------------------

Comments

These posts may also be useful for you

top 3 websites for downloading research papers for free

plot graph in gnuplot from csv and data file | knowledge of physics

projectile motion in fortran | relation of angle of projection and horizontal range

NEB physics exam numerical problems with solutions for grade 12 students

how to plot 3D and parametric graphs in gnuplot | three dimensional plot in gnuplot software

how to create GIF animation in gnuplot | animation using gnuplot software

NEB board exam maths solved problems with proper solutions - available for free | short answered mathematics problems solved group A

NEB grade 11 and 12 maths exam MCQ solved problems | NEB mathematics MCQ solved problems