`Letf(x)={{:(x+1,", "if xge0),(x-1,", "if xlt0):}".Then prove that" lim_(xto0) f(x)` does not exist.
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We have `f(x)={{:(x+1,", "if xge0),(x-1,", "if xlt0):}".` This function is piecewise function which changes its definition at x=0. So, to find limiting value of f(x) at x=0 we find LHL and RHL. To the left of zero, we have f(x) `=(x-1)` and to the right of zero, we have f(x) `=(x+1).` `LHL=underset(xto0)limf(x)=underset(hto0)limf(0-h)=underset(hto0)limf(-h)` `underset(hto0)lim((-h)-1)` -1 `RHL=underset(xto0^(+))limf(x)=underset(hto0)limf(0+h)=underset(hto0)limf(h)` `underset(hto0)lim(h+1)` `=1` Thus, LHL`ne`RHL. Therefore, `underset(xto0)limf(x)` does not exist.
LINEAR COMBINATION OF VECTORS, DEPENDENT AND INDEPENDENT VECTORS
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