Find lim h->0 [f(2+h)-f(2)]/h where f(x)=|x-5|-7
f(2+h) = |2+h-5|-7 = |h-3| -7
as h---> 0 this is 3-h-7 = -h -4
f(2) = |-3| -7 = 3-7 = -4
so
[ -h - 4 + 4 ] /h
= h/h
= 1
At x=2, x-5 < 0, so the slope at x=2 is just -1.
Ok, ok, algebraically,
f(2+h) = |2+h-5|-7 = |-3+h|-7
Since the value is negative, that will be 3-h-7 = -4-h
f(2) = |2-5|-7 = 3-7 = -4
-4-h - (-4) = -h
Now, divide by h and you have -h/h = -1
-h/h
or
-1
To compute the limit of the given expression as h approaches 0, we will substitute the value of f(x) into the expression and simplify it step by step.
First, let's find f(2+h) and f(2):
f(2+h) = |(2+h)-5| - 7
= |h-3| - 7
f(2) = |2-5| - 7
= |-3| - 7
= 3 - 7
= -4
Now, substitute the values back into the expression:
lim(h -> 0) [f(2+h) - f(2)] / h
= lim(h -> 0) [(|h-3| - 7) - (-4)] / h
= lim(h -> 0) [|h-3| - 3] / h
Since the limit involves an absolute value function, we need to consider the cases when h-3 is positive or negative.
When h-3 ≥ 0, |h-3| = h-3.
When h-3 < 0, |h-3| = -(h-3) = 3-h.
Splitting the limit into these two cases:
Case 1: h-3 ≥ 0
lim(h -> 0) [(h-3) - 3] / h
= lim(h -> 0) (h - 6) / h
= lim(h -> 0) h/h - 6/h
= lim(h -> 0) 1 - 6/h
= 1 - 0
= 1
Case 2: h-3 < 0
lim(h -> 0) [-(h-3) - 3] / h
= lim(h -> 0) (3 - h - 3) / h
= lim(h -> 0) (-h) / h
= lim(h -> 0) -1
= -1
Since the limits for the two cases are different (-1 and 1), the limit of the overall expression does not exist.
Therefore, lim(h -> 0) [f(2+h)-f(2)]/h = DNE (Does Not Exist).