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

a = 4

b = -2

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