This is defined when the value shown by the function at the points to the left of a point ‘a’ is the left-hand limit of the function at that point. It is denoted as lim x→a− f(x). This is equal to the value shown by it at the rightward of a point ‘a’ which shows the value of the function. This is the right-hand limit of the function at that point and is denoted as lim x→a+ f(x). Thus we can say that a function is defined when the left-hand limit is equal to the right-hand limit and which equals the value of the function at that point.
Properties of Limits
Lim x→a c=c, i.e limit of a constant function is that constant value only
The value of lim x→a x = a
Value of lim x→a cx + d = ca + d
Lim x→a xn = an, here n is a positive integer
Lim x→0− 1/xr = −∞, if r is odd, and
Lim x→0− 1/xr = +∞, if r is even
In case of evaluation of limits, the following identities are followed:
If lim x→a m(x) = α and lim x→a n(x) = β then and if both α and β exist, then the below rules are valid for finding out the limits
lim x→a [m(x) + n(x)] = lim x→a m(x) + lim x→a n(x) = α + β
lim x→a [m(x) - n(x)] = lim x→a m(x) - lim x→a n(x) = α - β
lim x→a [m(x) . n(x)] = lim x→a m(x) .lim x→a n(x) = α . β
lim x→a [m(x) / n(x)] = lim x→a m(x) /lim x→a n(x) = α / β If β≠0
Different Methods for the Evaluation of Limits
Direct Substitution Method: If limits lim x→a f(x) and lim x→a φ(x)/Ψ(x). If f(a) and φ(a)/Ψ(a) exist and are fixed real numbers and Ψ(a) ≠ 0, then we say that lim x→a f(x) = f(a) while lim x→a φ(x)/Ψ(x) = φ(a)/Ψ(a).
Factorization Method: If lim x→a f(x)/g(x). In substituting x = a, f(x)/g(x) reduces to the form 0/0, then (x – a) will be a factor of both f(x) and g(x). In this case, we have to first factorize f(x) and g(x) and then we can cancel out the common factor to evaluate or find the limit.
Rationalization Method: It is used in cases where either the numerator or the denominator or both involve expressions consisting of square roots and on substituting the value of x, the rational expression takes the form 0/0, ∞/∞.
Evaluation of Algebraic limits at infinity: If lim x→∞ 1/x = 0 and lim x→∞ 1/x2 = 0.Then lim x→∞ f(x) = lim y→0 f(1/y).
If f, g, h are three given continuous functions such that f < g < h in an open interval containing 'a' and also lim x→a f(x) = lim x→a h(x) = l. Then, to find lim x→a g(x), we use this squeezing method that says lim x→a g(x)=l
From the given figure given below, we have f ≤ g ≤ h.
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Question 1. Find the value of limit given below
limx→2 [x3 + 2x2 + 4x – 2].
limx→2 [x3 + 2x2 + 4x – 2] = limx→2 x3 + 2 limx→2 x2 + 4 limx→2 x – limx→2 2
= 23 + 2×22 + 4×2 − 2 = 22.
Question 2: Find the limit of limx→0 (sin 6x/tan x).
limx→0 (sin 6x / tan x) = limx→0 [(sin 6x / 6x) × (6x / tan x)] (multiplying and dividing by 6x)
or, limx→0 (sin 6x / tan x) = limx→0 (sin 6x / 6x) × 6 limx→0 (x / tan x )
= 1 × 6 = 6 (by using L-Hospital’s Rule).