Let the given function defined over a range as
f( x )={ 2x+3,x≤0 3( x+1 ),x>0
We have to find the values of the given function at limit x→0 and limit x→1 .
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 x→a f( x )=f( a )
So on solving the first expression for the value at x=0
lim x→ 0 − f( x )=( 2x+3 ) = lim x→0 ( 2x+3 ) =( 2⋅0+3 ) =3 (Left hand limit)
lim x→ 0 + f( x )=( 2x+3 ) = lim x→0 ( 2x+3 ) =( 2⋅0+3 ) =3 (Right hand limit)
We see from the above two expression that:
lim x→ 0 − f( x )= lim x→ 0 + f( x )= lim x→0 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 x→1 3( x+1 ) =( 3⋅1+3 ) =( 3+3 ) =6 (Left hand limit)
lim x→ 1 + f( x )=3( x+1 ) = lim x→1 3( x+1 ) =( 3⋅1+3 ) =( 3+3 ) =6 (Right hand limit)
Here, again we see that:
lim x→ 1 − f( x )= lim x→ 1 + f( x )= lim x→1 f( x )=6 (2)
Since, lim x→ a − f( x )= lim x→ a + f( x )= lim x→a f( x )
Hence, the limit exists.
Thus, from equations (1) and (2), the value of the given expressions is
lim x→0 f( x )=3
And
lim x→1 f( x )=6