Question 182519
<font face="Times New Roman" size="+2">


The *[tex \Large n^2] in the radicand increases without bound as <i>n</i> increases without bound because:


*[tex \LARGE \text{          }\math \lim_{x\to a} x^m = a^m] so


*[tex \LARGE \text{          }\math \lim_{x\to \infty} x^m = \infty]


Same thing for the *[tex \Large n]  in the radicand because *[tex \Large n = n^1]


The 1 in the radicand remains 1 regardless of the value of <i>n</i>, so we can say:



*[tex \LARGE \text{          }\math \lim_{x\to a} b = b] so


*[tex \LARGE \text{          }\math \lim_{x\to \infty} b = b]


But the limit of the sum is the sum of the limits, so if we say:


*[tex \LARGE \text{          }\math G(n) = n^2 + n + 1], then 


*[tex \LARGE \text{          }\math \lim_{n\to \infty} G(n) = \infty + \infty + 1 = \infty]


Now if we define *[tex \Large H(x) = sqrt(x) = x^{1 \over 2}], we can say:


*[tex \LARGE \text{          }\math \lim_{x\to \infty} H(x) = \infty] directly from


*[tex \LARGE \text{          }\math \lim_{x\to \infty} x^m = \infty]


So that means


*[tex \LARGE \text{          }\math \lim_{G(n)\to \infty} H(G(n)) = \infty]


The factor of 8:


*[tex \LARGE \text{          }\math \lim_{x\to \infty} b = b] so


*[tex \LARGE \text{          }\math \lim_{n\to \infty} 8 = 8]


The factor of *[tex \Large \pi]:


*[tex \LARGE \text{          }\math \lim_{x\to \infty} b = b] so


*[tex \LARGE \text{          }\math \lim_{n\to \infty} \pi = \pi]


And finally the factor of *[tex \Large n^2] outside of the radicand:


As previously shown:


*[tex \LARGE \text{          }\math \lim_{n\to \infty} n^2 = \infty]


The limit of the product is the product of the limits, so:


*[tex \LARGE \text{          }\math \lim_{n\to \infty} 8n^2\pi sqrt{n^2 + n + 1} = (\lim_{n\to \infty}8)(\lim_{n\to \infty}n^2)(\lim_{n\to \infty}\pi)(\lim_{n\to \infty}sqrt{n^2 + n + 1})\cr\cr\text{          } =(8)(\infty)(\pi)(\infty)=\infty]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>