Evaluate the left-and right-hand limits of the function defined by
`f(x)={1+x^2,if0lt=x<1 2-x ,ifx >1`
at `x=1.`
Also, show that `("lim")_(xvec1)f(x)`
does not exist
Text Solution
Verified by Experts
LHL of `f(x)` at x=1 is `underset(xto1^(-))limf(x)=underset(hto0)f(1-h)` `=underset(hto0)lim[1+(1-h)^(2)]` `=underset(hto0)lim(2-2h+h^(2))=2` RHL of f(x) at x=1 is `underset(xto1^(+))limf(x)=underset(hto0)limf(1+h)` `=underset(hto0)lim[2-(1+h)]` `underset(hto0)lim(1-h)=1` Clearly, `underset(xto1^(-))limf(x)neunderset(xto1^(+))f(x)` So, `underset(xto1)limf(x)` does not exist.
LINEAR COMBINATION OF VECTORS, DEPENDENT AND INDEPENDENT VECTORS
CENGAGE PUBLICATION|Exercise DPP 1.2|10 Videos
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