Some questions about limits (1 Viewer)

bawd

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Could someone please explain what these questions are asking for, and then how to do them?

Show that the following limits do not exist:

1. lim x --> 0 f(x) = { 0 when x < 0 , 1 when x > 0

2. lim x --> 0 f(x) = { x when x > 0 , x + 1 when x < 0

3. lim x --> 0 f(x) = { x^2 + a when x > 0 , 2 when x < 0

Thanks!
 

Trebla

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bawd said:
Could someone please explain what these questions are asking for, and then how to do them?

Show that the following limits do not exist:

1. lim x --> 0 f(x) = { 0 when x < 0 , 1 when x > 0

2. lim x --> 0 f(x) = { x when x > 0 , x + 1 when x < 0

3. lim x --> 0 f(x) = { x^2 + a when x > 0 , 2 when x < 0

Thanks!
lim x--> 0 f(x), means the value that f(x) takes as we creep very close to x = 0, but not actually reaching it. If a limit does not exist, then when we sub values close to x = 0 on both sides, they do not converge to the same number.
1) lim x --> 0 f(x) = { 0 when x < 0 , 1 when x > 0
So lim x --> 0+ f(x) = 1 (approaching x from positive end)
and lim x --> 0- f(x) = 0 (approaching x from negative end)
Since lim x --> 0+ f(x) =/= lim x --> 0- f(x), then the limit does not exist.

2) lim x --> 0 f(x) = { x when x > 0 , x + 1 when x < 0
So lim x --> 0+ f(x) = 0 (= x which approaches 0)
and lim x --> 0- f(x) = 1 (= x + 1 and x approaches 0)
Since lim x --> 0+ f(x) =/= lim x --> 0- f(x), then the limit does not exist.

3) lim x --> 0 f(x) = { x^2 + a when x > 0 , 2 when x < 0
So lim x --> 0+ f(x) = a (= x2 + a and x approaches 0)
and lim x --> 0- f(x) = 2
Since lim x --> 0+ f(x) =/= lim x --> 0- f(x), then the limit does not exist.
 

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