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

Find The Value Of A And B For Which The Limit Exists Both As X Approaches 1 And As X Approaches 2 class=

Sagot :

Answer:

a = 4

b = -2

Step-by-step explanation:

If the given function is continuous at x = 1

[tex]\lim_{x \to 1^{-}} f(x)=(x+1)[/tex]

                     [tex]=2[/tex]

[tex]\lim_{x \to 1^{+}} f(x)=ax+b[/tex]

                     [tex]=a+b[/tex]

[tex]\lim_{x \to 1} f(x)=ax+b[/tex]

                   [tex]=a+b[/tex]

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

[tex]\lim_{x \to 1^{-}} f(x)=\lim_{x \to 1^{+}} f(x)=\lim_{x \to 1} f(x)[/tex]

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

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

[tex]\lim_{x \to 2^{-}} f(x)=ax+b[/tex]

                     [tex]=2a+b[/tex]

[tex]\lim_{x \to 2^{+}} f(x)=3x[/tex]

                     [tex]=6[/tex]

[tex]\lim_{x \to 2} f(x)=3x[/tex]

                   [tex]=6[/tex]

Therefore, [tex]\lim_{x \to 2^{-}} f(x)=\lim_{x \to 2^{+}} f(x)=\lim_{x \to 2} f(x)[/tex]

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

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