![\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:
4/16
Step-by-step explanation:
Answer:

Step-by-step explanation:

Collecting like terms and simplifying yields

Answer:
Step-by-step explanation:
Solve each inequality one at a time:
5x ≤ 21 + 2x and
3x ≤ 21 so
x ≤ 7
The other inequality is
3x + 3 ≥ 24 and
3x ≥ 21 so
x ≥ 7
If x is less than or equal to 7 AND greater than or equal to 7, then the solution is all real numbers.