Consider sinc(x):where sinc(x)=sin (xpi)/xpi. and sinc(x) have max value i.e. u

Consider sinc(x):where sinc(x)=sin (xpi)/xpi.
and sinc(x) have max value i.e. unity at x=0 .now if
x=0.then pix =0 and sin (pix)=0 at x=0. then above
expression sin (xpi)/xpi.should be of 0/0 form i.e.
in-determined form. then how sinc(x) is one at x=0.
pi is constant.

Question Posted / tejas g

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Consider sinc(x):where sinc(x)=sin (xpi)/xpi. and sinc(x) have max value i.e. unity at x=0 .now if ..

Answer / gauravdeep singh

when x=0, sinc(x) becomes indeterminate i.e 0/0 form.Hence
apply L Hospitals rule i.e Lt x->0 {d/dx(sincx)}...which will
result in cos(xpi)..which is 1 fr x=0

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