![\bf \lim\limits_{x\to \infty}~\left( \cfrac{1}{8} \right)^x\implies \lim\limits_{x\to \infty}~\cfrac{1^x}{8^x}\\\\[-0.35em] ~\dotfill\\\\ \stackrel{x = 10}{\cfrac{1^{10}}{8^{10}}}\implies \cfrac{1}{8^{10}}~~,~~ \stackrel{x = 1000}{\cfrac{1^{1000}}{8^{1000}}}\implies \cfrac{1}{8^{1000}}~~,~~ \stackrel{x = 100000000}{\cfrac{1^{100000000}}{8^{100000000}}}\implies \cfrac{1}{8^{100000000}}~~,~~ ...](https://tex.z-dn.net/?f=%5Cbf%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D~%5Cleft%28%20%5Ccfrac%7B1%7D%7B8%7D%20%5Cright%29%5Ex%5Cimplies%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D~%5Ccfrac%7B1%5Ex%7D%7B8%5Ex%7D%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bx%20%3D%2010%7D%7B%5Ccfrac%7B1%5E%7B10%7D%7D%7B8%5E%7B10%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B10%7D%7D~~%2C~~%20%5Cstackrel%7Bx%20%3D%201000%7D%7B%5Ccfrac%7B1%5E%7B1000%7D%7D%7B8%5E%7B1000%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B1000%7D%7D~~%2C~~%20%5Cstackrel%7Bx%20%3D%20100000000%7D%7B%5Ccfrac%7B1%5E%7B100000000%7D%7D%7B8%5E%7B100000000%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B100000000%7D%7D~~%2C~~%20...)
now, if we look at the values as "x" races fast towards ∞, we can as you see above, use the values of 10, 1000, 100000000 and so on, as the value above oddly enough remains at 1, it could have been smaller but it's constantly 1 in this case, the value at the bottom is ever becoming a larger and larger denominator.
let's recall that the larger the denominator, the smaller the fraction, so the expression is ever going towards a tiny and tinier and really tinier fraction, a fraction that is ever approaching 0.
Answer:
Hi
Step-by-step explanation:
Hello, how' life dude?
Answer:
5.39 units
Step-by-step explanation:
Pythagorean theorem: 
Let point (-2,1) be C
Line segment AC would be 2 units long.
Line segment BC would be 5 units long.
Insert the length of the line segments in to the Pythagorean theorem

4 + 25 = 29
≅ 5.385
Answer:
1.9
Step-by-step explanation:
Use triginometry to find the side length. So cosθ=
. Rearrange the equation so it looks like 2cosθ=?. We are given θ (20 degrees) so 2cos(20degrees)=?.
Answer:a
Step-by-step explanation: