Solve this question.....
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f(0)=lim f(x)
x-o
= lim [(1+x)*1/3-1]÷x
x-0
This is a (0/0) form, so we apply L Hospital's rule and differentiate denominator and numerator for remove the(0/0) form
=lim [1/3(1+x)^-2/3]/1
x-0
=1/3
For the continuity of the given function
lim f(x)=f(0)
x-0
So f(0)=1/3
x-o
= lim [(1+x)*1/3-1]÷x
x-0
This is a (0/0) form, so we apply L Hospital's rule and differentiate denominator and numerator for remove the(0/0) form
=lim [1/3(1+x)^-2/3]/1
x-0
=1/3
For the continuity of the given function
lim f(x)=f(0)
x-0
So f(0)=1/3
deep1171:
can we do without L'hospital rule ??
if you can eliminate (0/0) form then you can do it but it's not easy
Okay thank you...
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