Hey, I'm studying for a test on Wednesday. Would I be able to cancel out the '^16' in this equation to get ((x+1)^16)/x^16 = 1?

No....we can't do that.....

We have

(x + 1)^16 / x^16  =  1          multiply bothsides by x^16

(x + 1)^16   =  x^16             subtract x^16 from both sides

(x + 1)^16  - x^16   = 0             we can factor this several times, as follows

[  ( x + 1)^8  + x^8 ]  [ (x + 1)^8 - x ^8]   = 0

The first factor won't  have a real solution.....so.....working with the second

[ (x + 1)^4  + x^4]  [ (x + 1)^4 - x^4]  = 0

Again, the first has no real solution.....factor the second

[ ( x+ 1)^2  + x^2] [ (x + 1)^2 - x^2]  = 0 

No real solution for the first.......factor the second one more time

[ ( x + 1) + x]  [ (x + 1) - x ]  = 0

The second has no real solution......set the first to 0

(x + 1) + x  = 0     subtract 1 from both sides

2x  = -1        divide both sides by 2

x  = -1/2        this is the only real solution    

The short answer is YES!

lim_(x->∞) (x + 1)^16/x^16 = 1

Hey, I'm studying for a test on Wednesday.

Would I be able to cancel out the "^16" in this equation to get lim_(x->∞) (x + 1)^16/x^16 = 1?

\(\begin{array}{|rcll|} \hline && \lim \limits_{x\to \infty} \dfrac{ (x + 1)^{16} } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \dfrac{ \binom{16}{0} x^{16} + \binom{16}{1} x^{15} + \binom{16}{2} x^{14} + \ldots + \binom{16}{15} x^1 + \binom{16}{16} } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \dfrac{ 1\cdot x^{16} + 16 \cdot x^{15} + 120 \cdot x^{14} + \ldots + 16\cdot x + 1 } {x^{16}} \\ &=& \lim \limits_{x\to \infty} \frac{ x^{16} } {x^{16}} + 16 \cdot \frac{ \cdot x^{15}} {x^{16}} + 120 \cdot \frac{ x^{14}} {x^{16}} + \ldots + 16\cdot \frac{ x } {x^{16}} + \frac{ 1 } {x^{16}} \\ &=& \lim \limits_{x\to \infty} 1 + \frac{ 16} {x} + \frac{ 120} {x^2} + \ldots + \frac{ 16 } {x^{15}} + \frac{ 1 } {x^{16}} \\ &=& 1 + 0 + 0 + \ldots + 0 + 0 \\ &=& 1 \\ \hline \end{array} \)