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प्रश्न
Examine the continuity of the function:
f(x) = `(log100 + log(0.01+x))/"3x"," for "x != 0 = 100/3 `for x = 0; at x = 0.
योग
उत्तर
for `x!= 0, f(x)= (log 100 + log(0.01+x))/(3x) = (log(1+100x))/(3x)`
For continuity at x = 0, f(0) = f(0-) = f(0 +)
`therefore lim_(x->0^-)f(x) = lim_(x->0^-)(log (1+100x))/(3x)= lim_(x->0^-)1/((1+100x)(-3)) (-100)=100/3`
`therefore lim_(x->0^+) f(x) = lim_(x->0^+)(log(1+100x))/(3x) = lim_(x->0^+)1/((1+100x)3) (100) = 100/3`
As f(0) = f(0-) = f(0 +) = `100/3`, the function is continuous at x= 0 .
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