Question

Determine the following limits, if it exists:
3) lim x->0 (cube rootx)-1 / (square rootx)-1
4) lim n->infinity n(n)!/(n+2)!

Answer 1

(n+2)!=(n+2)(n+1)*n!

Answer 2

so you should be able to write 4 without factorials

Answer 3

3 says
\[\lim_{x \rightarrow 0}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}\]?

Answer 4

if so try a sub

Answer 5

\[\text{ \to get rid of the ugly radicals \choose a sub such that its power } \\ \text{ is divisible by both 3 and 2 }\]
like 6
so we choose the following sub:
\[x=u^6 \]

Answer 6

solving that for u we see as x goes to 0 u also goes to 0

Answer 7

\[\lim_{u \rightarrow 0} \frac{u^\frac{6}{3}-1}{x^\frac{6}{2}-1}\]
say by by the ugly radicals

Answer 8

now you reduce the powers
and factor

Answer 9

and cancel

Answer 10

and plug in

Answer 11

can't you rationalize denominator then factor a difference of cubes

Answer 12

\[(\sqrt[3]{x}-1)(\sqrt{x}+1)=x^{\frac{1}{3}+\frac{1}{2}}+x^\frac{1}{3}-x^\frac{1}{2}-1 =x^\frac{5}{6}+x^\frac{1}{3}-x^\frac{1}{2}-1\]
if i'm not mistaken this is what you get for the numerator when trying to rationlize the denominator as is

Answer 13

I think the sub simplifies things

Answer 14

\[\lim_{u \rightarrow 0} \frac{u^\frac{6}{3}-1}{u^\frac{6}{2}-1}=\lim_{u \rightarrow 0}\frac{u^2-1}{u^3-1}=\lim_{u \rightarrow 0}\frac{(u-1)(u+1)}{(u-1)(u^2+u+1)}\]

Answer 15

so it will be 1?

Answer 16

yea (0+1)/(0^2+0+1)=1/1=1

Answer 17

thanks lots! but i dont understand the 4) question.

Answer 18

do you understand
that (n+2)!=(n+2)*(n+1)*n!

Answer 19

no

Answer 20

do you know what 5! equals?

Answer 21

120?

Answer 22

and how did you get that

Answer 23

graphing calculator...

Answer 24

I was looking for this
\[5!=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \]

Answer 25

oh yeah...

Answer 26

\[n!=n \cdot (n-1) \cdot (n-2) \cdot (n-3) \cdots 3 \cdot 2 \cdot 1 \]

Answer 27

okay...

Answer 28

\[5!=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \text{ or we can write \it as } \\ 5!=5 \cdot 4! \text{ or even } 5!=5 \cdot 4 \cdot 3! \]

Answer 29

\[(n+2)!=(n+2) \cdot (n+1) \cdot n! \]
I stopped at the n part here
because I have n! on top and bottom now

Answer 30

\[\lim_{n \rightarrow \infty}\frac{n \cdot n!}{(n+2)(n+1) \cdot n!}\]

Answer 31

\[\lim_{n \rightarrow \infty}\frac{n}{(n+2)(n+1)}\]

Answer 32

oh i see..
how would i sub in the infinity?

Answer 33

multiply the bottom out and use that
\[\lim_{x \rightarrow \pm \infty}\frac{1}{x^r}=0 \text{ where } r>0\]

Answer 34

\[\lim_{n \rightarrow \infty}\frac{n}{n^2+3n+2} =\lim_{n \rightarrow \infty}\frac{\frac{n}{n^2}}{\frac{n^2}{n^2}+\frac{3n}{n^2}+\frac{2}{n^2}}\]

Answer 35

I must go now
but you should be able to do this part now

Answer 36

thank you sooo much!