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$$\lim_{}\frac{5^{n-2}-3^{n+1}}{4^{n-1}+5^n} \Rightarrow \lim_{}\frac{\frac{5^{n-2}}{5^n}+\frac{3^{n+1}}{5^n}}{\frac{4^{n-1}}{5^n}+\frac{5^n}{5^n}} \Rightarrow \lim_{}\frac{5^{n-2-n}-\left(\frac{3}{5}\right)^{n+1-n}}{\left(\frac{4}{5}\right)^{n-1-n} + 1}$$

Is the last step valid? If not how do i solve the limit when n approaches infinite? Thanks in advance.

 

EDIT: Thanks for the answer.

Is this what you mean by split up: $$\lim_{}\frac{5^{n-2}}{4^{n-1}+5^n} - \lim_{}\frac{3^{n+1}}{4^{n-1}+5^n}$$ ?

Hyabak  Jan 9, 2015
 #1
avatar+93633 
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That last step looks pretty strange to me.  You can't do that with indices.  Do you have an answer?

 

suggestions

1)  Split it up into two separate limits.  It will be easier to work that way

2) do  the second one first

2a) Divide top and bottom by 5^n

2b) It will become  0/(0+1) = 0

 

3) Now looking at the first one

3a) split 5^(n-2) into  5^n/25   and take the 1/25 out the front.

3b) divide top and bottom by 5^n

3c) it becomes (1/25)(1/(0+1)) = 1/25

 

4)   1/25+0=1/25

 

You see if you can do the working :)

DO YOU UNDERSTAND WHAT I MEAN?  Let me know if you get stuck.  I will be around just a little longer. :)

Melody  Jan 9, 2015
 #2
avatar+93633 
0

Sorry - I did not ralise that you had answered.

It is better to put it in a new post so that it can be seen.

YES   that is exactly what I mean :)

Also, this is how you do the limit in LaTex

 

\displaystyle \lim_{x\rightarrow \infty}

$$\displaystyle \lim_{x\rightarrow \infty}$$

Melody  Jan 9, 2015

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