Find lim f(x) for f(x)=Ix-3I/x-3

x->3

not sure if i understand you when you say from the left and right?

from the left: -(x-3)/(x-3)=-1

from the right: (x-3)/(x-3)=1

To find the limit of f(x) as x approaches 3, we need to evaluate the function as x approaches 3 from both the left and the right side.

Let's start with the left side approach:

As x approaches 3 from the left side, we can substitute x = 3 - h into the function, where h > 0.

f(x) = Ix - 3I / x - 3

f(3-h) = I(3 - h) - 3I / (3 - h) - 3

To simplify the absolute value expression, note that when x < 3, Ix - 3I is equal to -(x - 3).

f(3-h) = -h / -h
= 1

Next, let's consider the right side approach:

As x approaches 3 from the right side, we can substitute x = 3 + h into the function, where h > 0.

f(x) = Ix - 3I / x - 3

f(3+h) = I(3 + h) - 3I / (3 + h) - 3

To simplify the absolute value expression, note that when x > 3, Ix - 3I is equal to (x - 3).

f(3+h) = (h) / h
= 1

Since the limit from both sides is the same (1), then we can conclude that the limit of f(x) as x approaches 3 is equal to 1.

Therefore, lim f(x) as x->3 is equal to 1.

To find the limit of the function f(x) as x approaches 3, we can evaluate the function for values of x that are very close to 3 from both sides.

First, let's consider values of x that are slightly less than 3. For these values, the expression inside the absolute value Ix-3I will be negative, so we need to consider the negation of that expression.

As x approaches 3 from the left side, x < 3. So we have:
f(x) = Ix-3I / (x-3)
= -(x-3) / (x-3)
= -1

Now, let's consider values of x that are slightly greater than 3. For these values, the expression inside the absolute value Ix-3I will be positive, so we can simply evaluate the expression as is.

As x approaches 3 from the right side, x > 3. So we have:
f(x) = Ix-3I / (x-3)
= (x-3) / (x-3)
= 1

Since the value of f(x) approaches -1 from the left side and 1 from the right side as x approaches 3, we can conclude that the limit of f(x) as x approaches 3 is -1.

Therefore, lim f(x) = -1
x->3