f(x)= {kx-3 if x is less than or = to 1

{x^2+k if x is greater than 1

find k such that lim f(x) (x to 1) exists

To find the value of k such that the limit of f(x) as x approaches 1 exists, we need to ensure that the left-hand limit (limit from the negative side) and the right-hand limit (limit from the positive side) are equal.

Let's start by finding the left-hand limit (LHL) and the right-hand limit (RHL) of f(x) as x approaches 1.

LHL:
As x approaches 1 from the left-hand side (x < 1), the function f(x) is defined as kx - 3. So we substitute x = 1 into the expression:
LHL = lim(x->1-) (kx - 3) = k(1) - 3 = k - 3

RHL:
As x approaches 1 from the right-hand side (x > 1), the function f(x) is defined as x^2 + k. So we substitute x = 1 into the expression:
RHL = lim(x->1+) (x^2 + k) = (1)^2 + k = 1 + k

To find the value of k such that the limit of f(x) as x approaches 1 exists, we need to set LHL equal to RHL and solve for k:

k - 3 = 1 + k

We can observe that k cancels out, resulting in a contradiction. Therefore, there is no value of k for which the limit of f(x) as x approaches 1 exists.