Solving Limit of \frac{(n+4)^{100} - (n+3)^{100}}{(n+2)^{100} - n^{100}}

[twoflower]

Hi all, I can't find limit of this one:

<br /> \lim \frac{(n + 4)^{100} - (n + 3)^{100}}{(n + 2)^{100} - n^{100}}<br />

I only got it to this point after I divided all expressions with n^100:

<br /> \lim \frac{ \left( 1 + \frac{4}{n} \right) ^{100} - \left( 1 + \frac{3}{n} \right) ^{100}}{ \left( 1 + \frac{2}{n} \right) ^{100} - 1}<br />

I only can see that every expression goes to 1 in infinity, but I can't figure the limit out of this, anyway...

Thank you for any suggestions. I would like to ask as well, what to do in cases like this - when I get \frac{0}{0} or \frac{\infty}{\infty} (and without l'Hospital).

Thank you.

 

[matt grime]

we can expand each bracket and write the quantity as:

\frac{100n^{99} + P(n)}{200n^{99} + Q(n)}

where P and Q are polynomials of degree 98. Divide by n^{99} top and bottom and take the limit.

 

[twoflower]

Thank you matt, but could you please show me the way you got the expression you posted? I can't see how to expand the brackets...Thank you much.

 

[matt grime]

You're doing limits but don't know how to expand brackets using the binomial theorem?

http://mathworld.wolfram.com/BinomialTheorem.html

 

[twoflower]

Actually, that was the first I tried, but I didn't see the possibility to write the sums as a sum of two polynoms, one of which will go to zero when divided with n^99. Now I have it. Thank you for your time matt.