lim as h approaches 0: (sqrt(4+h))- 2)/h

Question

Please help me understand how to do this problem.

Answer 1

There are at least two ways to do this problem.  I'll show you both of them.  

 

Method 1 (algebra):

 

Multiply the numerator and denominator by the conjugate of the numerator, √(4+h)+2, to obtain the following limit (which is equivalent to the given limit):

  

lim_h→0{h/[h(√(4+h)+2]} = lim_h→0(1/(√(4+h)+2)] = 1/4

---------------------------------------------------------------------------------

 

Method 2: (definition of derivative)

 

Let f(x) = √x.  Then f(4+h) = √(4+h) and f(4) = 2

 

So, the given limit is f'(4) = lim_h→0[(f(4+h)-f(4))/h]

 

   Now, since f'(x) = (1/2)x^-1/2 

                          = 1/(2√x), we have f'(4) = 1/(2√4) = 1/4

 

 

Answer 2

 

The other answers are perfectly correct.   Just want to point out you can also use L'Hospital's Rule.   This rule applies for indeterminate cases like limits of functions that look like 0/0, ∞/∞, and a few other cases.

 

The rule says, for indeterminate cases:

 

                     Lim {  f(x)/g(x) }  =  Lim  { f'(x) } / Lim {g'(x)}   

                     x->k                         x->k               x->k

 

This is a very powerful rule indeed!

 

Since the numerator  (4 + h)^1/2 - 2  —>0  as h —> 0   and the denominator obviously does, L'Hospital's Rule can be used.

 

Let f(h) = (4+h)^1/2 - 2

Let g(h) = h

 

 

f'(h) = ½(4 + h)^-1/2

g'(h) = 1

 

So    lim  { ((4+h)^1/2-2) / h } = lim  { ½(4 + h)^-1/2 } / lim { 1 }

       h—>0                               h—>0                         h—>0

 

                                             = lim { (1/2) (1/(4+h)^1/2) }      NOTE the exponent is positive 1/2 due to reciprocal 

                                               h—>0

 

                                              = (1/2) (1/2) 

                                              = 1/4