Ex 5.1, 9 Find all points of discontinuity of f, where f is defined by π(π₯)={β(π₯/|π₯| , ππ π₯<0@&β1 , ππ π₯β₯ 0)β€ Since we need to find continuity at of the function
We check continuity for different values of x
When x = 0
When x > 0
When x < 0
Case 1 : When x = 0
f(x) is continuous at π₯ =0
if L.H.L = R.H.L = π(0)
Since there are two different functions on the left & right of 0, we take LHL & RHL .
if limβ¬(xβ0^β ) π(π₯)=limβ¬(xβ0^+ ) " " π(π₯)= π(0)
And,
f(0) = β1
LHL at x β 0
limβ¬(xβ0^β ) f(x) = limβ¬(hβ0) f(0 β h)
= limβ¬(hβ0) f(βh)
= limβ¬(hβ0) (ββ)/|ββ|
= limβ¬(hβ0) (ββ)/β
= limβ¬(hβ0) β1
= β1
RHL at x β 0
limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h)
= limβ¬(hβ0) f(h)
= limβ¬(hβ0) β1
= β1
Hence, L.H.L = R.H.L = π(0)
β΄ f is continuous at x = β3
Case 2 : When x < 0
For x < 0,
f(x) = π₯/(|π₯|)
f(x) = π₯/((βπ₯))
f(x) = β1
Since this constant
It is continuous
β΄ f(x) is continuous for x < 0
(As x < 0, x is negative)
Case 3 : When x > 0
For x > 0,
f(x) = β1
Since this constant
It is continuous
β΄ f(x) is continuous for x > 0
β΄ f is continuous for all real numbers
Thus, f is continuous for πβ R
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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