InterviewSolution
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InterviewSolution
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निम्नलिखित फलनों की सांतत्यता की जाँच x = 0 पर कीजिए - (a) `f(x) = {{:((|sin x|)/(x)",",x ne 0),(1",",x = 0):}` (b) `f(x) = {{:(x "sin"(1)/(x)",",x ne 0),(0",",x = 0):}` |
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Answer» (a) यहाँ `f(x)=(|sin x|)/(x),` (i) `f(0)=1` (ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)` `=underset(h rarr 0)(lim)(|sin (0 + h)|)/((0+h))," "[because x = 0 + h ne 0]` `=underset(h rarr 0)(lim)(|sin h|)/(h)` `=underset(h rarr 0)(lim)(sin h)/(h)` `= 1" "[because underset(theta rarr 0)(lim)(sin theta)/(theta)=1]` (iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)` `=underset(h rarr 0)(lim)(|sin (0-h)|)/((0-h))` `=underset(h rarr 0)(lim)(|sin (-h)|)/(-h)` `=underset(h rarr 0)(lim)(|-sin h|)/(-h)," "[because sin (-theta)=sin theta]` `=underset(h rarr 0)(lim)(sin h)/(-h)` `=-1" "[because underset(theta rarr 0)(lim)(sin theta)d/(theta)=1]` `therefore" "underset(x rarr 0^(-))(lim)f(x) ne underset(x rarr 0^(+))(lim)f(x)=f(0)` अत: `f(x), x = 0` पर असंतत है । (b) यहाँ `f(x)=x "sin"(1)/(x)` (i) `f(0)=0` (ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x)=underset(h rarr 0)(lim)f(0+h)` `=underset(h rarr 0)(lim)(0+h)"sin"(1)/((0+h))` `=underset(h rarr 0)(lim)h sin((1)/(h))` `= 0 xx` परिमिति राशि, [`because "sin"(1)/(h)` का मान -1 और +1 के मध्य स्थित होता है] = 0 (iii) L.H.L. `=underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)` `=underset(h rarr 0)(lim)(0-h)sin((1)/(0-h))` `=underset(h rarr 0)(lim)(-h)sin(-(1)/(h))` `=underset(h rarr 0)(lim)(-h)(-sin((1)/(h)))" "[because sin(-theta)=-sin theta]` `=underset(h rarr 0)(lim)hsin((1)/(h))` `=0 xx` अपरिमित राशि, [`because "sin"(1)/(h)` का मान -1 और +1 के मध्य स्थित होता है] = 0 `therefore" "underset(x rarr 0^(+))(lim)f(x)=underset(x rarr 0^(-))(lim)f(x)=f(0)` अत: `f(x), x = 0` पर संतत है । |
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