Recent content by kingstrick

got it! thanks!

since |a| < 1 then lim |a| = |a| < 1 and L = lim |a| * L then L - L*Lim |a| = 0 which implies L(1-Lim|a|)=0 therefore L = 0

but doesn't that show lim |a|^(n+1) = lim |a| * lim |a|^n L = lim |a| * L 1 = lim |a| ??

lim |a|^(n+1) = L

It should have the same limit

lim |a|^(n+1) = aL < L since a < 1

I am unsure. I remember from an earlier class that the limit of x^n is equal to n * lim ln x so i tried extending it to calculate the limits like you suggested: lim of a^(n+1) = (n+1) * lim ln a [1] lim (a*a^n) = lim a * lim a^n = lim a * n * lim ln a. [2] Then i divided both sides ([1]...

i am still missing something, cause now i am getting the lim to be e which i know is incorrect. (n+1) * lim ln |a|= Lim (|a| * |a|^n) = lim |a| * n lim ln |a| (n+1) = n lim |a| (1+1/n) = lim |a| (1+1/n)^n = (lim |a|)^n e = lim |a|^n

Homework Statement given |a| < 1, show that the limit of |a|^n goes to 0 as n goes to infinity. Homework Equations The Attempt at a Solution let |a|<1 and n>0 (n is a natural number, a is a real number) then |a^n| < 1^n then |a|^n < 1 then 1/n * |a|^n <=...

I guess I am asking if this is a iff statement. That if 1/g(c) exist implies that (1/g(x)) is continuous at c. Meaning that if g(x) is continuous thrn g(c) exists and given it is not 0 then 1/g(c) exists then 1/g(x) is continuous at c.

If it is given that g(c) exists and does not equal zero, Doesnt that mean 1/g(c) exists?

If |g(x)| > |g(c)|/2 then 2/|g(c)| >1/|g(x)| Allowing me to state 2/|g(c)g(c)| > 1/|g(c)g(x)|

**corrected

I am not sure how to handle the fraction So now I have: (\frac{1}{g(x)g(c)})|g(c)||f(x)-f(c)|+|f(c)||g(x)-g(c)| given that |x-c|<\delta then (\frac{1}{g(x)g(c)})|g(c)||f(x)-f(c)|+|f(c)||g(x)-g(c)|=(\frac{1}{g(x)g(c)})|g(c)||x-c|+|f(c)||x-c|<\epsilon so to find a good delta-epsilon i would...

Can I say |f(x) - f(c)| <ϵepsilon/g(c) since g(c) is a value and not equal to zero?