Explain why the function is discontinuous at the given number a = 1 ?

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Answer 1 · 2018-06-04T02:03:09

The limit of the first piece as x approaches 1 from the negative side is:

#lim_(x to 1^-) e^x = e ~~ 2.718#

We do need to take a limit for the second piece; we can just evaluate it at 1:

#1^2 = 1#

The function is discontinuous because the two pieces do not have the same y value at 1;

Answer 2 · 2018-06-04T02:25:10

See below

#f(x)# is a piecewise function defined above.

To show that #f(x)# is discontinuous at a given point we can examine the behaviour of the function at that point.

In this case #f(1) = 1^2 = 1# because #f(x) = x^2: x in [1, +oo)#

Now let's consider #f(1-delta)# for some arbitrarily small #delta >0#

In this case #f(1-delta) = e^(1-delta)# because #(1-delta) < 1 forall delta >0#

So as #delta -> 0# then # f(1^-) -> e#

Since #f(1) != f(1^-) -> f(x)# is discontinuous at #x=1#

[NB: #f(1^-)# means the limit of #f(x)# as #x -> 1# from below.]