If `f(x)={{:(x-1", "xge1),(2x^(2)-2", "xlt1):},g(x)={{:(x+1", "xgt0),(-x^(2)+1", "xle0):

Correct Answer - `(0)`
`underset(xto0^(+))limf(g(h(x)))=f(g(0^(+)))=f(1^(+))=0`
`underset(xto0^(-))limf(g(h(x)))=f(g(0^(+)))=f(1^(+))=0`
Hence, `underset(xto0)limf(g(h(x)))=0`

If `f(x)={{:(x-1", "xge1),(2x^(2)-2", "xlt1):},g(x)={{:(x+1", "xgt0),(-x^(2)+1", "xle0):}," and "` `h(x)=|x|," then "lim_(xto0) f(g(h(x)))" is"`______

Please log in or register to add a comment.

1 Answer

Please log in or register to add a comment.

Find MCQs & Mock Test

Related questions

Categories