The left hand derivative of $$f(x)=$$[x]$$sin$$⁡$$($$π$$x)$$ at $$x=k$$,k an integer, is

Question

The left hand derivative of $$f(x)=$$[x]$$sin$$⁡$$($$π$$x)$$  at  $$x=k$$,k an integer, is

Infinity Learn · Accepted Answer

The left hand limit L at $$x=k$$ is  given  by                                     $$L=limh$$→$$0k$$−hsin$$π$$k−h−h                                          $$=limh$$→$$0k$$−$$1sin$$$$π$$kcos$$π$$h−$$cos$$$$π$$ksin$$π$$has k−h  when  k is an is an integer, is equal to (k$$−$$1).Now,   $$sin$$$$π$$$$k=0$$  and  $$cos$$$$π$$$$k=$$−$$1k$$∴                              $$L=limh$$→$$0(k$$−$$1)0$$−−$$1ksin$$$$π$$h−h                                     $$=$$ (k$$−$$1).π.Hence (1) is the correct answer.

$The left hand derivative of f (x) = [x] \sin (π x) a t x = k, k an integer, is$

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The left hand derivative of f(x)=[x]sin⁡(πx) at x=k,k an integer, is

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$The left hand derivative of f (x) = [x] \sin (π x) a t x = k, k an integer, is$