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Limits
Graphical Interpretation of Limit of a Function
Left and Right Limit
One Sided Limits
Algebra of Limits
Related Resources
Limits is an extremely important topic of calculus. It is also important because it lays the groundwork for various other topics like continuity and differentiability. Each and every notion of calculus can be considered to be a limit in one sense or the other. The concept of limits has also resulted in various other branches of calculus.
Consider a function f(x) = (x^{2 }- a^{2})/(x – a)
The function f(x) is defined at all points except at x = a, because at x = a, f(x) = 0/0 which is indeterminate in the language of mathematics, hence we want to know what value does f(x) approach as x approaches a. The limiting behaviour of the function in the neighborhood of x = a when |x - a| is small, is called the limit of the function when x approaches 'a' and we write this as lim _{x→a} f(x).
Let lim _{x→a} f(x) = l. It would simply mean that when we approach the point x = a from the values which are just greater than or just smaller than x = a, f(x) would have a tendency to move closer to take value 'l' This is same as saying, "difference between f(x) and l can be made as small as we like by suitably choosing x in the neighborhood of x = a". Mathematically, we write this as, lim _{x→a}f(x) = l, which is equivalent to saying that |f(x) – l| < ε for all x such that 0 < |x - a| < δ and ε depends on δ where ε and δ are sufficiently small positive numbers.
It should be clear that the limit of f(x) at x = a would exist if and only if, f(x) is well defined in the neighborhood of x = a (not necessarily at x = a) and has a unique behaviour in the neighborhood of x = a.
Normally students have the perception that limit should be a finite number. But it is not really so. It is quite possible that f(x) has infinite limit as x → a. If lim _{x}_{→}_{a} f(x) = ∞, it would simply mean that function has tendency to assume very large positive values in the neighborhood of x = a, for example lim _{x}_{→}_{0} 1/|x| = ∞.
As discussed above, in case of limits, we are not concerned with the fact that whether the function is defined at the particular point say ‘a’ or not, but we are more concerned with the value attained by the function at the point ‘a’. We have further clarified this concept with the help of diagrams given below. The first figure depicts the situation where f(c) = L. The second figure depicts the condition where the function f is not defined at c. The last case, i.e. the third figure depicts the situation where though the function f is defined at ‘c’ but f(c) ≠ L. However, in each case, as x approaches or tends to c, the function f(x) approches L.
Let y = f(x) be a given function, and x = a is the point under consideration. Left tendency of f(x) at x = a is called its left limit and right tendency is called its right limit.
Left tendency (left limit) is denoted by f(a - 0) or f(a^{-}) and right tendency (right limit) is denoted by f(a + 0) or f(a^{+}) and are written as
f(a – 0) = lim _{x→a} f(a – h)
f(a + 0) = lim _{x→a} f(a + h), where ‘h’ is a small positive number.
Thus for the existence of the limit of f(x) at x = a, it is necessary and sufficient that f(a – 0) = f(a + 0), if these are finite or f(a – 0) and f(a + 0) both should be either + ∞ or - ∞.
For the existence of the limit at x = a, f(x) need not be defined at x = a. However if f(a) exists, limit need not exist or even if it exists then it need not be equal to f(a).
For what values of m does the lim _{x→2} f(x) exist when
f(x) = mx – 3, when x < 2
= x/m , when x ≥ 2 ?
lim _{x→2- }f(x) = lim _{x→2-} f (mx – 3) = 2m - 3;
lim _{x→2+} f(x) = lim _{x→2+} x/2 = 2/m
lim _{x→2} f(x) exists when lim _{x→2-} f(x) = lim _{x→2+} f(x)
⇒ 2m - 3 = 2/m
⇒ 2m^{2} - 3m -2 = 0
⇒ m = -1/2, 2.
Let f(x) be a function defined on domain D and let ‘a’ be a point such that every neighborhood of ‘a’ contains infinitely many points of D. A real number ‘l’ is called the left hand limit of f(x) at x = a iff for every ∈ > 0 there exists a δ > 0 such that a – δ < x < a ⇒ |f(x) – l| < ∈.
In such a case we write lim _{x→a-} f(x) = l.
Thus, lim_{ x→a-} f(x) = l if f(x) tends to l as x tends to a from the left hand side.
Similarly, a real number ‘l’ is said to be a right hand limit of f(x) at x = a iff for every ∈ > 0, there exists a δ > 0 such that
a < x < a + δ ⇒ | f(x) – l | < ∈ and we write it as lim _{x→a+} f(x) = l*.
Hence, l* is said to be the right hand limit of f(x) at x = a iff f(x) tends to l* as x tends to ‘a’ from the right hand side.
Let the function f(x) be defined in x ∈ [a, b]. At times, we might have to calculate lim _{x→b} f(x) or lim _{x→a} f(x).
In such a case, lim _{x→a} f(x) = lim _{x→a+} f(x) = R.H.L at x = a, as there is no left neighborhood of x = a.
Similarly, lim _{x→b} f(x) or lim _{x→b-} f(x) = L.H.L at x = b as there is no left neighborhood of x = b.
Eg: f(x) = cos^{-1} x
lim _{x→1} f(x) = lim _{x→1-} cos^{-1} x = 0.
lim _{x→ -1} f(x) = lim _{x→ -1}+ cos^{-1} x = π.
For more on limits, please view the below listed video
Suppose lim _{x→a} f(x) = α and lim _{x→a} g(x) = β then, if both α and β exist, then we can define the following rules:
lim x→a [f(x) + g(x)] = lim x→a f(x) + lim x→a g(x) = α + β
lim x→a [f(x) - g(x)] = lim x→a f(x) + lim x→a g(x) = α + β
lim x→a [f(x) . g(x)] = lim x→a f(x) + lim x→a g(x) = α + β
lim x→a fog(x) = f(lim x→a g(x)) = f(β), only if f is continuous at g(x) = β
lim x→a |f(x)| = |lim x→a f(x)| = |α|
lim x→a [f(x)]g(x) = lim x→a [f(x)]lim x→a g(x) = αβ
If lim x→a f(x) = + ∞ or -∞ then
If f(x) ≤ g(x) for every x in the neighborhood of a, then lim x→a f(x) ≤ lim x→a g(x).
These results have been discussed in detail in the section on “Algebra of Limits”.
If the value of a function at any point acquires one of the forms 0/0; 0 × ∞; ∞/∞; ∞ – ∞; 0^{0}; 1^{∞}, ∞^{0} then the function is said to be indeterminate at that point. (The first of these is said to be standard indeterminate form)
It is again important to note that one is not allowed to manipulate the function i.e. we cannot write (x^{2 }- a^{2})/(x - a) = (x + a)(x – a)/((x - a)) = (x + a) because this we can do only if (x- a) ≠ 0 or x ≠ a. At x = a, this manipulation is not valid due to such restriction, as the function, locating a break in the value of f(x) at x = a is almost impossible. So we can interpret that value of the function at point infinitesimally close to a exists and is finite.
Graph of y = (x^{2 }- a^{2})/(x – a)(x ≠ a) is shown in figure given below circle at point (a, 2a) means that the point is excluded from the graph of the function.
It is for this reason that we define limits and also get rid of the manipulation or simplification constraint of function. So the moment we write
lim _{x→a} (x^{2 }- a^{2})/(x – a) we can write lim _{x→a} (x + a)(x – a))/(x – a) = lim _{x→a}(x + a) = 2a
(Remember lim _{x→a} (x – a) ≠ 0)
Now continuing with same argument and considering the same example we can infer whether the limit exists at a point where the function is not defined
The condition for the existence of limit of a function at x = a is Limit of f(x) as x approaches 'a' From Left = Limits of f(x) as x approaches 'a' From Right
i.e. Left Hand Limit (LHL) at (x = a) = Right Hand Limit (RHL) at (x = a) or mathematically
lim _{x→a-} f(x) = lim _{x→a+} f(x)
⇒ For figure given below, Limit exists, L is the value of limit whether x approaches 'a' from left or right, even though f(a) is not defined. For the second figure, Limit exists, L is the limit when x approaches 'a' from left and L^{2 }is limit as x approaches 'a' from right.
Graphically one can infer that, if there is no sudden change or break of value of f(x) slightly left and slightly right of point x = a, it means existence of limit of f(x) at x = a.
let f: R → R be such that f(1) = 3 and f’(1) = 6. The find the value of lim _{x→0} [f(1 + x)/ f(1)]^{1/x}. (IIT JEE 2002)
Let y = [f(1 + x)/ f(1)]^{1/x}
So, log y = 1/x [log f(1 + x) – log f(1)]
So, lim _{x→0} log y = lim _{x→0 }[1/f (1+x) . f’(1+x)]
= f’(1)/ f(1)
= 6/3
log (lim _{x→0} y) = 2
lim _{x→0} y = e^{2}.
^{___________________________________________________________________________}
lim _{x→0 }sin(π cos^{2}x)/x^{2} equals (IIT JEE 2001)
(a) -π (b) π
(c) π/2 (d) 1
lim _{x→0} sin (π cos^{2}x)/x^{2 }= lim _{x→0} sin (π – π sin^{2}x)/x^{2 }
^{ }=^{ }sin (π sin^{2}x) / (π sin^{2}x) . (π sin^{2}x) / πx^{2 }. π
= 1 . 1 . π = π.
Hence, the correct option is (b).
Q1. If lim _{x}_{→}_{a} f(x) = L, then
(a) the function f must be defined at ‘a’.
(b) the function f can’t be defined at the point ‘a’.
(c) the funtcion f may or may not be defined at the point ‘a’.
(d) none of these
Q2. While solving a question on limits, (x^{2 }- a^{2})/(x – a) = (x + a)
(a) can be done provided (x-a) ≠ 0 or x ≠ a
(b) is correct, as we have just cancelled out the common terms
(c) cannot be done like this at all in case of limits
Q3. By definition, lim _{x}_{→}_{a }f(x) = l, which is equivalent to saying that
(a) |f(x) – l| < ε for all x such that 0 < |x - a| < δ and ε depends on δ where ε and δ are sufficiently small positive numbers.
(b) |f(x) – l| > ε for all x such that 0 < |x - a| < δ and ε depends on δ where ε and δ are sufficiently small positive numbers.
(c) |f(x) – l| < ε for all x such that 0 < |x - a| < δ and δ depends on ε where ε and δ are sufficiently small positive numbers.
(d) |f(x) – l| < ε for all x such that 0 < |x - a| < δ and ε depends on δ where ε and δ are sufficiently small negative numbers.
Q4. Limit of a function
(a) has to be a finite number
(b) may be finite or infinite
(c) has to be an infinite number
Q5. If f(a) exists, then
(a) the limit need not exist, but if it exists, it must be equal to f(a).
(b) the limit must exist and it must be equal to f(a).
(c) the limit of the function f must exist at the point a but necessarily equal to f(a).
(d) limit need not exist and even if it exists, it need not be equal to f(a).
Q1.
Q2.
Q3.
Q4.
Q5.
(c)
(a)
(b)
(d)
You may wish to refer Continuity.
Click here to refer the most Useful Books of Mathematics.
For getting an idea of the type of questions asked, refer the previous year papers.
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