InterviewSolution
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InterviewSolution
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निम्नलिखित फलनों के सांतत्यता की जाँच x = 0 पर कीजिए - (a) `f(x)={{:(e^(1//x)",",x ne 0),(0",",x = 0):}` `f(x)={{:(e^(1//x)/(1+e^(1//x))",",x ne 0),(0",",x = 0):}` (c) `f(x) = {{:(e^(1//x-1)/(e^(1//x)+1)",",x ne 0),(0",",x = 0):}` |
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Answer» (a) यहाँ `f(x)=e^(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)e^(1//(0+h))," "[because x = 0 + h ne 0]` `=underset(h rarr 0)(lim)e^(1//h)` `=e^(1//10)=e^(oo)=e^(oo)` (iii) L.H.L. `underset(x rarr 0^(-))(lim)f(x)=underset(h rarr 0)(lim)f(0-h)` `=underset(h rarr 0)(lim)e^(1//(0-h))," "[because x = 0 - h ne 0]` `=underset(h rarr 0)(lim)e^(-1//h)` `=e^(-oo)=0," "[because e^(-oo)=0]` `therefore" "underset(x rarr 0^(+))(lim)f(x) ne underset(x rarr 0^(-))(lim)f(x) = f(0)` अत: `f(x), x = 0` पर संतत नहीं है । (b) यहाँ `f(x) = (e^(1//x))/(1+e^(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)(e^(1//(0+h)))/(1+e^(1//(0+h)))" "[because x = 0 + h ne 0]` `=underset(h rarr 0)(lim)(e^(1//h))/(1+e^(1//h))` `=underset(h rarr 0)(lim)(e^(1//h))/(e^(1//h)[e^(-1//h)+1])` `=underset(h rarr 0)(lim)(1)/(e^(-1//h)+1)` `= (1)/(e^(-oo)+1)=(1)/(0+1)=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)(e^(1//(0-h)))/(1+e^(1//(0-h)))," "[because x = 0 - h ne 0]` `=underset(h rarr 0)(lim)(e^(-1//h))/(1+e^(-1//h))` `=(e^(-oo))/(1+e^(-oo))=(0)/(1+0)=0," "[because e^(-oo)=0]` `therefore" "underset(x rarr 0^(+))(lim)f(x) ne underset(x rarr 0^(-))(lim)f(x) = f(0)` अत: f(x) बिन्दु x = 0 पर संतत नहीं है । (c) यहाँ `f(x) = (e^((1)/(x))-1)/(e^((1)/(x))+1)` (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)(e^(1//0+h)-1)/(e^(1//0+h)+1)," "[because x = 0 + h ne 0]` `=underset(h rarr 0)(lim)(e^(1//h)-1)/(e^(1//h)+1)` `=underset(h rarr 0)(lim)(e^(1//h)[1-e^(-1//h)])/(e^(1//h)[1 + e^(-1//h)])` `=underset(h rarr 0)(lim)(1-e^(-1//h))/(1+e^(-1//h))` `=(1-e^(-oo))/(1+e^(-oo))=(1-0)/(1+0)=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)(e^(1//(0-h))-1)/(e^(1//0-h)+1)," "[because x = 0 - h ne 0]` `=underset(h rarr 0)(lim)(e^(-1//h)-1)/(e^(-1//h)+1)` `=(e^(-oo)-1)/(e^(-oo)+1)=(0-1)/(0+1)=-1` `therefore" "underset(x rarr 0^(+))(lim)f(x) ne underset(x rarr 0^(-))(lim)f(x) ne f(0)` अत: `f(x), x = 0` पर संतत नहीं है । |
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