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If `f(x)={{:((x-|x|)/(x)","xne0),(2", "x=0):},`show that `lim_(xto0) f(x)` does not exist. |
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Answer» L.H.L of `f(x)` at `x=0` is `underset(xto0)limf(x)=underset(hto0)limf(0-h)=underset(hto0)lim(-h-|-h)/((-h))` `=underset(hto0)lim(-h-h)/(-h)=underset(hto0)lim(-2h)/(-h)=underset(hto0)lim2=2` R.H.L of `f(x)` at `x=0` is `underset(hto0)limf(x)=underset(hto0)limf(0+h)=underset(hto0)lim(h-|h)/((h))` `underset(hto0)lim(h-h)/(h)=underset(hto0)lim0/h=underset(hto0)lim0=0` Clearly, `underset(xto0^(-))limf(x)neunderset(xto0^(+))limf(x)` So, `underset(xto0^(-))limf(x)` does not exist. |
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