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)⋅g(x)]=limx→mf(x)⋅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.
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)2⋅limx→01ex
= 1. 1e0, ∴ e0=1.
= 1. 1
= 1. //
2.) Solve: limθ→0cosec θ-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-cosθθ2⋅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 △OXM ≤ Area of sector AXM ≤ Area of △OXY
..........................(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.
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