Aspire Faculty ID #16314 · Topic: NIMCET 2010 · Just now
NIMCET 2010

If $ f(x)= \begin{cases} x \sin\left(\frac{1}{x}\right), & x \ne 0 \\ 0, & x = 0 \end{cases} $ then

Solution

Solution: lim_{x→0} x·sin(1/x) = 0 ⇒ f is continuous at 0. f'(0) = lim_{x→0} [x sin(1/x)]/x = sin(1/x) But sin(1/x) has no limit as x→0⁺ or x→0⁻. ⇒ Both f'(0+) and f'(0-) do not exist.

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