Taking the limit of this problem, need a little help?

Question

Lim (1/h)*[ (1/SQRT(9+h)) - (1/3) ]
h-->0

Answer 1

lim [ (1/sqrt(9 + h) - (1/3) ] / h
h -> 0

First thing you should do is change this complex fraction into a simple fraction, by multiplying top and bottom by 3sqrt(9 + h). This will eliminate all fractions within the fraction. It will also leave stuff behind:

lim [ (3) - sqrt(9 + h) ] / [h ( 3sqrt(9 + h) )]
h -> 0

Now, rationalize the numerator by multiplying top and bottom by the conjugate of the top. The conjugate of [3 - sqrt(9 + h)] is
[3 + sqrt(9 + h)]. Remember that multiplying a binomial by its conjugate will result in a difference of squares; (a - b) times (a + b) is equal to (a^2 - b^2). So we have

lim [ (3)^2 - (sqrt(9 + h))^2 ] / { [h ( 3sqrt(9 + h) )][3 + sqrt(9 + h)]}
h -> 0

Simplify.

lim [9 - (9 + h)] / { [h ( 3sqrt(9 + h) )][3 + sqrt(9 + h)] }
h -> 0

We can simplify the top now. Notice that we're just left with -h.

lim [-h] / { [h ( 3sqrt(9 + h) )][3 + sqrt(9 + h)] }
h -> 0

Now, notice how there's an h on the top and the bottom. We can cancel these.

lim (-1) / { [( 3sqrt(9 + h) )][3 + sqrt(9 + h)] }
h -> 0

And now we can safely plug in 0 for every instance of h.

(-1) / { [( 3sqrt(9 + 0) )][3 + sqrt(9 + 0)] }

(-1) / [ (3sqrt(9)) (3 + sqrt(9)) ]

(-1) / [ 3(3) (3 + 3) ]

(-1) / [9(6)]

-1/54

Our answer is -1/54

Answer 2

The answer is zero.

Lim of (1/SQRT(9+h)) - (1/3) as h -> 0 = 0
0*anything = 0, so even theough 1/h -> infinity, infinity*0 = 0