Differential equationsDetermine whether f'(0) exists.
9 months ago
Determine whether f'(0) exists. f(x)=x^(2) sin 1/x if x is not equal to 0, 0 if x=0
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Best Answer
channel7katyperryconcert Staff answered 9 months ago
Recall that: f'(a)= lim_(x -> a) (f(x)-f(a))/(x-a) f'(0)= lim_(x -> 0) (x^(2) sin ((1)/(x))-0)/(x-0) f'(0)= lim_(x -> 0)x sin ((1)/(x)) Squeeze theorem If h(x) <= f(x) <= g(x) And lim_(x -> a)h(x)=L And lim_(x -> a)g(x)=L Then lim_(x -> a)f(x)=L Since -1 <= sin ((1)/(x)) <= 1 Therefore -x <= x sin ((1)/(x)) <= x Since lim_(x -> 0)-x=0 And lim_(x -> 0)x=0 By Squeeze Theorem, We have lim_(x -> 0)x sin ((1)/(x))=0 Result: f'(0)=0
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