How do you find the limit #lim_(h->0)(sqrt(1+h)-1)/h# ?

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Answer 1 · 2018-05-19T13:25:26

#\frac{1}{2}#

The limit presents an undefined form #0/0#. In this case, you may use de l'hospital theorem, that states

#lim \frac {f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}#

The derivative of the numerator is

#\frac{1}{2sqrt(1+h)}#

While the derivative of the denominator is simply #1#.

So,

#\lim_{x\to 0} \frac{f'(x)}{g'(x)} = \lim_{x\to 0} \frac{\frac{1}{2sqrt(1+h)}}{1} =\lim_{x\to 0} \frac{1}{2sqrt(1+h)}#

And thus simply

#\frac{1}{2sqrt(1)}=\frac{1}{2}#

Answer 2 · 2018-05-19T15:37:53

# = 1/2 #

If you are unaware of l'hopitals rule...

Use:

#(1+x)^n = 1 + nx + (n(n-1))/(2!) x^2 + ... #

#=> (1 + h)^(1/2) = 1 + 1/2h - 1/8 h^2 + ... #

#=> lim_( h to 0) ((1 + 1/2 h - 1/8h^2 + ...) - 1 )/ h #

#=> lim_( h to 0) ( 1/2 h - 1/8h^2 + ... )/ h #

#=> lim_( h to 0) ( 1/2 - 1/8 h + ... ) #

# = 1/2 #