Question

3. Consider the piecewise defined function:

f(x)=ax + 5 if x< 5
f(x)=2ᵃ+³-6 if x=5
f(x)=x²-25/ax+5a if x > 5

(a) Find all values of a, if any, for which
Lim x→5 f(x) exists.
(b) Find all values of a, if any, for which f is
continuous at x = 5.

Answer 1

To find the values of a for which lim x→5 f(x) exists, evaluate the limits of each piece of the function as x approaches 5. The limit exists if a is any real number except for 0.

Step-by-step explanation:

To find the values of a for which lim x→5 f(x) exists, we need to evaluate the limits of each piece of the function as x approaches 5.

For the first piece of the function, f(x) = ax + 5 if x < 5, the limit can be found by substituting 5 into the equation: lim x→5 (ax + 5) = a(5) + 5 = 5a + 5.

For the second piece of the function, f(x) = 2^(a+3) - 6 if x = 5, the limit can be found by substituting 5 into the equation: lim x→5 (2^(a+3)-6) = 2^(a+3)-6.

For the third piece of the function, f(x) = (x^2-25)/(ax+5a) if x > 5, the limit can be found by substituting 5 into the equation: lim x→5 (x^2-25)/(ax+5a) = (5^2-25)/(a(5)+5a) = 0/(10a) = 0.

So, the limit of the function as x approaches 5 exists if a is any real number except for 0.