InterviewSolution
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InterviewSolution
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Show that `lim_(xto0) (e^(1//x)-1)/(e^(1//x)+1)` does not exist. |
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Answer» Let `f(x)=(e^(1//x)-1)/(e^(1//x)+1)` L.H.L of `f(x) " at "x=0` is `underset(xto0^(-))limf(x)=underset(hto0)lim(0-h)=underset(hto0)lim(e^(-1//h)-1)/(e^(-1//h)+1)` `underset(hto0^(-))lim(((1)/(e^(1//h))-1)/((1)/(e^(1//h))+1))=-1` `[becausehto0implies1/htoooimpliese^(1//h)toooimplies(1)/(e^(1//h))to0]` R.H.L. of `f(x)` at `x=0` is `underset(xto0)limf(x)=underset(hto0)limf(0+h)=underset(hto0)lim(e^(1//h)-1)/(e^(1//h)+1)` `=underset(hto0)lim((1-(1)/(e^(1//h)))/(1+(1)/(e^(h))))" "`[Dividing Nr and Dr by `e^(1//h`)] `=(1-0)/(1+0)=1` Clearly, `underset(xto0^(-))limf(x)neunderset(xto0^(+))limf(x )` Hence, `underset(xto0)limf(x)` does not exist. |
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