Question

find the value of "a" and "b" for which the limit exists both as x approaches 1 and as x approaches 2:

Answer 1

Answer:

a = 4

b = -2

Step-by-step explanation:

If the given function is continuous at x = 1

$\lim_{x \to 1^{-}} f(x)=(x+1)$

                     $=2$

$\lim_{x \to 1^{+}} f(x)=ax+b$

                     $=a+b$

$\lim_{x \to 1} f(x)=ax+b$

                   $=a+b$

And for the continuity of the function at x = 1,

$\lim_{x \to 1^{-}} f(x)=\lim_{x \to 1^{+}} f(x)=\lim_{x \to 1} f(x)$

Therefore, (a + b) = 2 -------(1)

If the function 'f' is continuous at x = 2,

$\lim_{x \to 2^{-}} f(x)=ax+b$

                     $=2a+b$

$\lim_{x \to 2^{+}} f(x)=3x$

                     $=6$

$\lim_{x \to 2} f(x)=3x$

                   $=6$

Therefore, $\lim_{x \to 2^{-}} f(x)=\lim_{x \to 2^{+}} f(x)=\lim_{x \to 2} f(x)$

2a + b = 6 -----(2)

Subtract equation (1) from (2),

(2a + b) - (a + b) = 6 - 2

a = 4

From equation (1),

4 + b = 2

b = -2