F(x)=|x-5| evaluate the following limits
lim x->5- ( f(x)-f(5))/(x-5)
lim x->5+ ( f(x)-f(5))/(x-5)
To evaluate these limits, we need to substitute the given expressions into the function and then simplify them.
1. lim x->5- ( f(x)-f(5))/(x-5):
First, let's substitute the values into the function f(x):
f(x) = |x-5|
f(5) = |5-5|
When x approaches 5 from the left (represented by the "-"), we have x values that are slightly smaller than 5. However, since the function is an absolute value function, it doesn't matter if x is slightly smaller or larger than 5; the output will always be the same. So, f(x) = |x-5| is equal to |5-5|, which equals 0.
Now, let's substitute these values into the limit expression:
lim x->5- ( f(x)-f(5))/(x-5) = lim x->5- ( |x-5| - |5-5|)/(x-5) = lim x->5- ( |x-5| - 0)/(x-5)
Since the denominator (x-5) approaches 0 as x approaches 5 from the left side, we have an indeterminate form 0/0. To resolve this, we can simplify the numerator by recognizing that |x-5| is equal to the distance between x and 5.
lim x->5- ( f(x)-f(5))/(x-5) = lim x->5- ( |x-5|/(x-5) = lim x->5- ( 1 )
Hence, the limit of the expression as x approaches 5 from the left side is equal to 1.
2. lim x->5+ ( f(x)-f(5))/(x-5):
Following the same logic, when x approaches 5 from the right side (represented by "+"), we have values of x that are slightly greater than 5. Again, since the function is an absolute value function, the output will always be the same regardless of whether x is slightly smaller or larger than 5. So, f(x) = |x-5| is equal to |5-5|, which equals 0.
Now, let's substitute these values into the limit expression:
lim x->5+ ( f(x)-f(5))/(x-5) = lim x->5+ ( |x-5| - |5-5|)/(x-5) = lim x->5+ ( |x-5| - 0)/(x-5)
Similar to the previous case, we have an indeterminate form of 0/0. We simplify the numerator using the definition of absolute value as the distance between x and 5.
lim x->5+ ( f(x)-f(5))/(x-5) = lim x->5+ ( |x-5|/(x-5) = lim x->5+ ( 1 )
Thus, the limit of the expression as x approaches 5 from the right side is also equal to 1.