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Question

23. Find,g/(x) and,,(s), where2x +3, xs03(x+) >0

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Solution

Let the given function defined over a range as

f( x )={ 2x+3,x0 3( x+1 ),x>0

We have to find the values of the given function at limit x0 and limit x1 .

There are two different expressions for a given function defined at

x greater than or equal to 0 and other is for all positive values of x .

We need to take two arbitrary values of x around the given domain for solving the function.

Let those values of x be 0 and 1.

Now, we need to find the left hand limit and right hand limit of individual expression for two values of x .

From the definition of limits, we know that:

lim xa f( x )=f( a )

So on solving the first expression for the value at x=0

lim x 0 f( x )=( 2x+3 ) = lim x0 ( 2x+3 ) =( 20+3 ) =3 (Left hand limit)

lim x 0 + f( x )=( 2x+3 ) = lim x0 ( 2x+3 ) =( 20+3 ) =3 (Right hand limit)

We see from the above two expression that:

lim x 0 f( x )= lim x 0 + f( x )= lim x0 f( x )=3 (1)

Again on solving the second expression for the value at x=1

lim x 1 f( x )=3( x+1 ) = lim x1 3( x+1 ) =( 31+3 ) =( 3+3 ) =6 (Left hand limit)

lim x 1 + f( x )=3( x+1 ) = lim x1 3( x+1 ) =( 31+3 ) =( 3+3 ) =6 (Right hand limit)

Here, again we see that:

lim x 1 f( x )= lim x 1 + f( x )= lim x1 f( x )=6 (2)

Since, lim x a f( x )= lim x a + f( x )= lim xa f( x )

Hence, the limit exists.

Thus, from equations (1) and (2), the value of the given expressions is

lim x0 f( x )=3

And

lim x1 f( x )=6


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