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Question

Examine the continuity of the function:
f(x)=log100+log(0.01+x)3x, for x0=1003
for x=0; at x=0.

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Solution

For x0, f(x)=log100+log(0.01+x)3x=log(1+100x)3x

For continuity at x=0, f(0)=f(0)=f(0+)

limx0f(x)=limx0log(1+100x)3x=limx01(1+100x)(3)(100)=1003

(Using L'Hospital's rule)

limx0+f(x)=limx0+log(1+100x)3x=limx0+1(1+100x)3(100)=1003

(Using L'Hospital's rule)

As f(0)=f(0)=f(0+)=1003, the function is continuous at x=0.

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