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).
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
=
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?
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
factor (x+h)^.5 ?
I do not know how
Somehow you have to know that sqrt(1+h) ---> 1 + .5 h +.....
Perhaps someone else knows how
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
To evaluate the limit lim h→0 (sqrt(1+h) - 1) / h without using L'Hopital's Rule, you can try manipulating the expression algebraically to simplify it. Here's an approach:
First, let's simplify the numerator:
sqrt(1+h) - 1
To get rid of the square root, you can multiply both the numerator and denominator by the conjugate of the numerator:
(sqrt(1+h) - 1) * (sqrt(1+h) + 1) / (sqrt(1+h) + 1)
This simplifies to:
(1+h - 1) / (sqrt(1+h) + 1)
Now, cancel out the h terms from the numerator:
h / (sqrt(1+h) + 1)
Next, we can simplify the denominator by multiplying both the numerator and denominator by its conjugate:
h * (sqrt(1+h) - 1) / ((sqrt(1+h) + 1) * (sqrt(1+h) - 1))
This simplifies to:
h * (sqrt(1+h) - 1) / (1+h - 1)
Further simplification:
h * (sqrt(1+h) - 1) / h
Finally, the h terms cancel out, and you're left with:
(sqrt(1+h) - 1)
Now, as h approaches 0, the entire expression approaches:
(sqrt(1+0) - 1) = (sqrt(1) - 1) = 1 - 1 = 0
Therefore, the value of the limit is 0, and you have evaluated it without using L'Hopital's Rule.