calculus (limits)

lim h>0 sqrt(1+h)-1/h

not sure how to factor this; not allowed to use L'Hopital's Rule. (that isn't taught at my school until Calc II & I'm in Calc I).

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1. I bet you mean

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

One way:
I guess you know how to take derivatives
d/dx (x)^.5 = .5 (x^-.5)
(of course right there you can say
x^-.5 at x = 1 so your result is
.5/1 = .5
so
d (x^.5) = .5 x^-.5 dx
so
d (sqrt (x)) dx = .5 dx/sqrt(x)

d sqrt(1+h) dx = .5 dx/sqrt(1+h)
if h -->0
d sqrt(1+h)dx = .5 sqrt(1)h/sqrt(1)
= .5 h
so sqrt(1+h) --> 1 + .5 h
sqrt(1+h)-1 > .5 h
divide by h and get
.5

series for sqrt(x) for x approximately 1 is

=

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2. I do know how to take derivatives however the course I'm in hasn't taught them yet (long story) so I can't use derivatives in finding the solution; I have to factor.

So perhaps a better question is: how do I factor the above expression in a way that will allow me to find the limit?

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3. Taylor series for f(x + a ) for x approximately 1 is

f(x)= f(1) + f'(1)(x-a) + f''(x-a)^2/2 ...
for f(x) = sqrt (x)
f(1) = (1)^.5 = 1
f'(1) = .5 (1)^-.5 = .5
f"(x) = -.25(1)^-1.5 etc = -.25

f(x+h) = 1 + .5*h -.25 h^2 /2 etc
subtravt 1
.5h -.125 h^2 tc
divide by h and let h-->0
.5 - .125 h --> .5

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4. factor (x+h)^.5 ?

I do not know how
Somehow you have to know that sqrt(1+h) ---> 1 + .5 h +.....

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5. Perhaps someone else knows how

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6. rationalize the numerator, that is, multiply top and bottom by (√(1+h) + 1)/(√(1+h) + 1)

lim (√(1+h) - 1)/h as h --->0
= lim (√(1+h) - 1)/h (√(1+h) + 1)/(√(1+h) + 1)
= lim (1+h -1)/[h( (√(1+h) + 1)/h)]
= lim 1/( (√(1+h) + 1)
= 1/(1+1) = 1/2

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