Class 9 – Limits
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Introduction: a) Suppose we are travelling from Kanpur to Lucknow by a train which will reach Lucknow at 8.00 a.m. As the time gets closer and closer to 8.00 a.m., the distance of the train from Lucknow gets closer and closer to zero (assuming the train is running on time). If we consider time as our independent variable, denoted by and distance as a function of time, say f (t), then we see that f (t) approaches zero as t approaches 0. We can say that the limit of f (t) is zero as t tends to zero. b) Let a regular polygon be inscribed in a circle of given radius. We notice the following points from geometry. i) The area of the polygon cannot be greater than the area of the circle however large the number of sides may be. ii) As the number of sides of the polygon increases indefinitely, the area of the polygon continually approaches the area of the circle. 
iii) Ultimately the difference between the area of the circle and the area of the polygon can be made as small as we please by sufficiently increasing the number of sides of the polygon. This is expressed by saying that the limit of the area of the polygon inscribed in a circle, as the number of sides increases indefinitely (or approaches infinity), is the area of the circle. Meaning of x → a: Let x be a variable and ‘a’ be a constant. If x assumes values nearer and nearer to ‘a’, then we say that ‘x tends to a’ or ‘x approaches a’ and is written as ‘x ® a’. By x ® a, we mean x _ a and x may approach ‘a’ from left or right, which is explained in the examples given below. i) Consider the function In this case y is not defined for x = 1. ii) When x approaches 1 through values less than 1 written as. 

1.1 
1.01 
1.001… 
1.00000001 
0.9 
0.99 
0.999 
… 
0.9999999 

2.1 
2.01 
2.001… 
2.00000001 
1.9 
1.99 
1.999 
… 
1.9999999 
i) If the left hand limit of a function is not equal to the right hand limit of the function, then the limit does not exist.
ii) A limit equal to infinity does not imply that the limit does not exist.
Example: The limit of a chord of a circle passing through a fixed point Q and a variable point P as P approaches Q is the tangent to the circle at P.
Solution: PQ is chord of a circle, when P approaches Q along the circle, then the chord becomes the tangents to thecircle at P.
Theorems on limits: The following results are of fundamental importance and are accepted here without proof. Let f and g be two functions such that Then
The above results can be stated asunder:
i) The limits of the sum of two functionsis equal to the sum of their limits,
ii) The limits of the difference of two functions is equal to the difference of their limits.
iii) The limits of the product of two functions is equal to the product of their limits.
iv) The limit of quotient of two functionsis equal to the quotient of their limits provided the limit of the divisor is not zero.
The above results hold good only if land m both exist.
Algebraic Limits: Under this category all those functions which are not trigonometric, inverse trigonometric, exponential and logarithmic functions.
We generally evaluate algebraic limits by using the following five methods.
i) Direct substitution
ii) Factorization
iii) Rationalization
iv) Using Formula
v) Using type.
i) Direct substitution: Directly substitute the value in thegiven expression and if we get a finite number, then that finite number is the limit of the given expression.
Example: Evaluate
Solution: .
Examples:
1.
2.
ii) Factorization: If assumes an indeterminate form when x = a , then there exists a common factorfor . We remove the common factors, and use the substitution method to find the limit.
Example: Evaluate:
Solution:
Example: Evaluate:
Solution:
Example: Evaluate:
Solution:
iii) Rationalization: In functions which involves squareroots, rationalization of either numerator or denominator will facilitate thework.
Example: Evaluate:
Solution:
Example: Evaluate:
Solution:
Similarly,
Example: Evaluate:
Solution: In this case, we rationalize both thenumerator and the denominator.
Given limits
Theorem:.
Proof:
Example: Evaluate .
Solution:
Example:If then find the value of n.
Solution:
Example:Evaluate
Solution:
Example:
Solution:
To find:
i) Divide the numerator and denominator ofthe fraction by the highest power of x present in the fraction.
ii) Use the idea of
Example: Evaluate
Solution:
Example: Evaluate
Solution:Put
Aliter:
As
A function will have a limiting value only if its right hand limit equals to its left hand limit.
Example: If
Solution: