4. Let f(x)= ) x^2+1,&xleqslant 1 x+2,&xgt 1f(x) does not exist.

Question

Answer 1

To show that the limit of the piecewise function

as

approaches 1 does not exist, we need to check the left-hand limit and the right-hand limit separately.The left-hand limit is the limit of

as

approaches 1 from the left. In this case,

for

. Evaluating the left-hand limit, we have:

The right-hand limit is the limit of

as

approaches 1 from the right. In this case,

for

. Evaluating the right-hand limit, we have:

Since the left-hand limit and the right-hand limit are not equal, the limit of

as

approaches 1 does not exist.

Soru