We say that the left-hand limit of f(x) as x approaches x_{0} (or the limit of f(x) as x approaches from the left) is equal to l_{1} if we can make the values of f(x) arbitrarily close to l_{1} by taking x to be sufficiently close to x_{0} and less than x_{0}. It is symbolically written as
f(x_{0}^{-}) = lim x ->x_{0}^{- }f(x) = l_{1}
We say that the right-hand limit of f(x) as x approaches x_{0} (or the limit of f(x) as x approaches from the right) is equal to l_{2} if we can make the values of f(x) arbitrarily close to l_{2} by taking x to be sufficiently close to x_{0} and greater than x_{0}. It is symbolically written as
f(x_{0}^{+}) = lim x ->x_{0}^{+ }f(x) = l_{2}
From the above discussions we conclude that
lim x->x_{0} f(x) = L exists if the following hold :
(i) lim x->x_{0}+ f(x) exists,
(ii) lim x->x_{0}- f(x) exists, and
(iii) lim x->x_{0}+ f(x) = lim x->x_{0}- f(x) = L
When we get different values for f(x) as x_{0} approaches from left and from right, we may say that the function does not exist.
The picture given below will illustrate the concept.
There is no function to the left of x_{0}. Hence the function is not defined.
Example 1 :
Use the graph to find the limits (if it exists). If the limit does not exist, explain why?
lim _{x->3} (4 - x)
Solution :
f(x) = (4 - x) lim _{x->3}- f(x) = 4 - 3 = 1 |
f(x) = (4 - x) lim _{x->3}+ f(x) = 4 - 3 = 1 |
lim _{x->3 }f(x) = 4 - 3 = 1
lim _{x->3}- f(x) = lim _{x->3}+ f(x) = lim _{x->3} f(x)
The function is defined at x -> 3. Hence the required limit is 1.
Example 2 :
Use the graph to find the limits (if it exists). If the limit does not exist, explain why?
lim _{x->1} (x^{2} + 2)
Solution :
f(x) = (x^{2} + 2) lim _{x->1}- f(x) = 1^{2} + 2 = 3 |
f(x) = (x^{2} + 2) lim _{x->1}+ f(x) = 1^{2} + 2 = 3 |
lim _{x->1 }f(x) = x^{2} + 2 = 1^{2} + 2 = 3
lim _{x->1}^{-} f(x) = lim _{x->1}+ f(x) = lim _{x->1} f(x)
The function is defined at x -> 1. Hence the required limit is 3
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