Posted by John on .
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).

calculus (limits) 
Damon,
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
= 
calculus (limits) 
John,
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? 
calculus (limits) 
Damon,
Taylor series for f(x + a ) for x approximately 1 is
f(x)= f(1) + f'(1)(xa) + f''(xa)^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 
calculus (limits) 
Damon,
factor (x+h)^.5 ?
I do not know how
Somehow you have to know that sqrt(1+h) > 1 + .5 h +..... 
John needs analysis help 
Damon,
Perhaps someone else knows how

calculus (limits) 
Reiny,
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