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2007-02-06 00:51:13 · 4 answers · asked by Kevin K 4 in Science & Mathematics Mathematics

sorry thats wrong the end should be f(0) is nt equal to 1. ie if you fill 0 into the equation the answer cant be 1

2007-02-06 00:53:00 · update #1

4 answers

one such function is (sinx)/x
limit of the function as x approaches 0 is 1,this can be proved, refer any book on calculus for the proof.
but at x=0 function is not defined.

2007-02-06 00:58:32 · answer #1 · answered by san 3 · 0 0

Let f(x)=x+1 for x not equal to 0 and set f(0)=0.

2007-02-06 01:06:00 · answer #2 · answered by mathematician 7 · 0 0

Think of a function with a jump discontinuity at 0.

2007-02-06 00:54:57 · answer #3 · answered by Anonymous · 0 0

needless to say, any a no longer equivalent to 0 will paintings. If we'd go with to circumvent the three factors, any b no longer equivalent to 0 will paintings too. all of us understand this simply by fact for the functionality to bypass by way of those 3 factors might might desire to have a a million/2-era or finished era equivalent to 3. If it does, it won't bypass by way of (0,0) which it would desire to do for x = 0.

2016-12-17 10:33:35 · answer #4 · answered by ? 4 · 0 0

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