# Need help finding the limit as h approaches 0 for f(a)= [(-2)] / [(a+1)(a+h+1)]

hala718 | Certified Educator

f(a) = [(-2/(a+1)(a+h+1)]

To find the limit of f(a) as h --> 0

We will substitute with h=0:

==> lim f(a) =  (-2/(a+1)(a+0+a)

= (-2/(a+1)(a+1)]

= -2/(a+1)^2

Then lim f(a) as h--> 0 = -2/(a+1)^2

william1941 | Student

The solution for f(a) = -2/[(a+1)(a+h+1)] for limit h --> 0. can be found by merely substituting h=0. If the result is defined, that is the required solution.

-2/[(a+1)(a+h+1)]

substitute h=0

=>-2/[(a+1)(a+0+1)]

=>-2/[(a+1)(a+1)]

=>-2/(a+1)^2

Now, -2/(a+1)^2 is defined.

Therefore : -2/[(a+1)(a+h+1)] for limit h --> 0 = -2/(a+1)^2

neela | Student

To find the limit of f(a) = {-2/[(a+1)(a+h+1)} as h --> 0.

Solution:

Put h = 0 in f(a) = -2/[(a+1)(a+0+1)]

f(a) = -2/[(a+1)(a+1)].

f(a) = -2/(a+1)^2.

Also,  for any h = 0+ ,  f(a) = -2/(a+1)(a+1) = -2/(a+1)^2.

For any h = 0- , f(a) = -2/(a+1)^2.

Therefore, Lt  f(a) =  lt-2/[(a+1)(a+h-1)] as h - -> 0 is  -2/(a+1)^2.