Calculus insanity!?

Question

Let f(x)=x^2 - 5

compute the following limit:

lim as h->0 ((f(h+5) - f(5)) / (h)

and save me from insanity :D

Answer 1

f(h + 5) = h^2 + 10h + 25
f(h + 5) - f(5) = h^2 + 10h
(f(h + 5) - f(5))/h = h + 10

As h gets very small, h + 10 gets closer and closer to 10, so the limit is 10.

Answer 2

Okay, so first just plug your function into that thing:

lim h-> 0 ((h+5)^2 - (5^2 - 5))/ h

So, as h -> 0 (h+5)^2 goes to 5^2= 25
then as h-> 0 - (25-5) goes to -20
BUT then as that h on the bottom goes to zero, things DO get crazy! you pretty much get 5/approaching 0

which means the limit is infinity, because any positive number is infinitely larger than zero, and any negative number is infinitely smaller (that would approach - infinity)

hopefully that made sense..
basically, the limit is infinity, but next time just try to do each term at a time and remember #/0 = infinity and infinity/# = 0

Answer 3

you want to calculate derivative at x=5
df/dx=2x at x=5 it is equal to10
and calculus is not insane