Calculating Limits at Horizontal Infinity

[airforceone]

Hi,

I'm trying to find this limit:

I tried dividing the top and bottom by the highest power of the denominator (n^2), but I somehow got 0 for the bottom. Is there a way to simplify this problem further, rather than just factoring the top and bottom?

The answer is 1.

Thanks!

 

[BigGlenntheHeavy]

\(\displaystyle \lim_{n\to\infty}\frac{(n+1)^{1/2}+(n+1)^{1/2}*n^{2}}{n^{1/2}+n^{1/2}*(n+1)^{2}}\)

\(\displaystyle = \ \lim_{n\to\infty}\frac{(n+1)^{1/2}(1+n^{2})}{n^{1/2}[1+(n+1)^{2}]}\)

\(\displaystyle = \ \lim_{n\to\infty}\frac{(n+1)^{1/2}(n^{2}+1)}{n^{1/2}(n^{2}+2n+2)}\)

\(\displaystyle = \ \lim_{n\to\infty}\frac{n^{1/2}*n^{2}}{n^{1/2}*n^{2}} \ = \ 1\)

\(\displaystyle Note: \ As \ n \ approaches \ infinity, \ 1, \ 2n, \ and \ 2 \ become \ superfluous.\)

 

[JuicyBurger]

BigGlenntheHeavy is correct... You only need to divide the coefficients of the highest order term... in this case The coefficients of top and bottom are all 1, and they each have the same order, therefore the limit is 1.

 

[airforceone]

Okay thanks guys!