# Evaluate the limit..

• December 8th 2009, 10:37 PM
nameck
Evaluate the limit..
1. The problem statement, all variables and given/known data

lim.......... (1 + 1/n^2)(7/n+1)
n->infinity

3. The attempt at a solution

lim.......... (1 + 1/n^2)(7/n+1)
n->infinity

= (7/n + 1) + (7/n^3 + n^2)

bring out 7 because constant..

7lim.......... (1/n + 1) + (1/n^3 + n^2)
n->infinity
.
.
.
.
.
7lim.......... (n^2 + n + 1) / (n^3 + 2n^2 +n)
n->infinity

limit does not exist.. am i correct?
• December 8th 2009, 10:45 PM
acc100jt
$\displaystyle \lim_{n\rightarrow\infty}\left(1+\frac{1}{n^{2}}\r ight)\left(\frac{7}{n}+1\right)=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n^{2}}\right)\lim_{n\right arrow\infty}\left(\frac{7}{n}+1\right)
$

$=\left(\lim_{n\rightarrow\infty}1+\lim_{n\rightarr ow\infty}\frac{1}{n^{2}}\right)\left(\lim_{n\right arrow\infty}\frac{7}{n}+\lim_{n\rightarrow\infty}1 \right)=(1+0)(0+1)=1$
• December 8th 2009, 10:51 PM
RockHard
Quote:

Originally Posted by nameck
1. The problem statement, all variables and given/known data

lim.......... (1 + 1/n^2)(7/n+1)
n->infinity

3. The attempt at a solution

lim.......... (1 + 1/n^2)(7/n+1)
n->infinity

= (7/n + 1) + (7/n^3 + n^2)

Note that as the limit goes to infinity that
$\frac{1}{n^2}$ and $\frac{7}{n} \to\0$

bring out 7 because constant..

7lim.......... (1/n + 1) + (1/n^3 + n^2)
n->infinity
.
.
.
.
.
7lim.......... (n^2 + n + 1) / (n^3 + 2n^2 +n)
n->infinity

limit does not exist.. am i correct?

First let me ask that this does not represent a series? as you can perform a variety of test on different series to determine convergence, as the standard variable used in a series to n.

$\lim_{n\to\infty}(1+\frac{1}{n^2})(\frac{7}{n}+1)$

Note that: $\frac{1}{n^2}$ and $\frac{7}{n} \to 0$

as we goes towards infinity leaving us with 1 * 1 = 1 the limit is 1