Question

EXERCISE 5.1

Prove that the function f(x) = 5x -- 3 is continuous at x = 0, at x=- 3 and at x = 5.
1 at Y-3​

Answer 1

Step-by-step explanation:

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Answer 2

$\bigstar \underline{\underline{\sf \bf To\ Prove:-}}$

• f(x) = 5x -- 3 is continuous at x = 0, at x=- 3 and at x = 5.

$\bigstar \underline{\underline{\sf \bf Proof:-}}$

We need to know:

A function f is continous at x = a if ,

• f(a) is defined

• $\sf \lim_{x\to a} f(x) \ exists$

• $\sf \lim_{x \to a} f(x)=f(a)$

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Case-1 :-

At x= 0 ,

LHS :

$:\implies \sf \lim_{x \to o} f(x)= \lim_{x\to 0}( 5x-3)\\\\:\implies \sf 5(0)-3= \bold{-3}.$

RHS :

$:\implies \sf f(0) = 5(0)-3 \\\\:\implies \sf 0-3 = \bold{-3}.$

$\\\\\therefore \sf \lim_{x\to 0} f(x)=f(0)$

LHS = RHS

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Case-2 :-

At x = -3 ,

LHS :

$:\implies \sf \lim_{x \to -3} f(x)\\\\:\implies \sf \lim_{x \to -3}(5x-3)\\\\:\implies \sf 5(-3)-3=\bold{-18}.$

RHS :

$:\implies \sf f(-3)=5(-3)-3\\\\:\implies \sf -15-3= \bold{-18}.$

$\therefore \sf \lim_{x \to -3} f(x)=f(-3)$

LHS = RHS

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Case-3 :-

At x = 5 ,

LHS :

$:\implies \sf \lim_{x \to 5} f(x)= \lim_{x \to 5} (5x-3)\\\\:\implies \sf 5(5)-3 =\bold{22}.$

RHS :

$:\implies \sf f(5)=5(5)-3\\\\:\implies \sf 25-3=\bold{22}.$

$\therefore \sf \lim_{x \to 5 } f(x)=f(5)$

LHS=RHS

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Hence, The function is continuous at x=0,-3,5.

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HOPE IT HELPS !