# Limit evaluation

• Nov 28th 2010, 05:57 AM
azarue
Limit evaluation
Hi forum members, I've attached 2 limits I'm struggling to evaluate. could you give me a lead on it ? thanks in advance.
• Nov 28th 2010, 06:20 AM
elemental
For the first one, you can find the limit of each term and sum them.
Since the degree of the numerator is less than the denominator's in each case, each term approaches $\displaystyle 0$. Hence, the limit is...
For the second one, it would be useful to rewrite it as $\displaystyle (3n^2 + 20n + 13)$ to the power $\displaystyle 1/n$. The exponent here approaches $\displaystyle 0$, and any positive number to the power $\displaystyle 0$ is...
• Nov 28th 2010, 06:46 AM
Defunkt
Quote:

Originally Posted by elemental
For the first one, you can find the limit of each term and sum them.
Since the degree of the numerator is less than the denominator's in each case, each term approaches $\displaystyle 0$. Hence, the limit is...
For the second one, it would be useful to rewrite it as $\displaystyle (3n^2 + 20n + 13)$ to the power $\displaystyle 1/n$. The exponent here approaches $\displaystyle 0$, and any positive number to the power $\displaystyle 0$ is...

The first one is wrong and the second one is misleading:

In the first one, the number of summands depends on n, which tends to infinity - that is why the limit of the sum does not equal the sum of the limits (only true when there is a fixed amount of summands!)

Instead, try evaluating a lower bound and an upper bound for the sum and then use the sandwhich theorem. I'll show you how to do the lower bound and you do the upper bound:
The smallest summand is $\displaystyle \frac{n}{n^2+n}$, therefore, if we denote the sum by $\displaystyle S_n$, we have that $\displaystyle \displaystyle S_n \ge n \cdot \frac{n}{n^2+n} = \frac{n^2}{n^2+n} = \frac{n}{n+1} \to 1$ as $\displaystyle n \to \infty$. Try working the upper bound in a similar fashion.

For the second one, you can evaluate the expression inside the root as follows: $\displaystyle \displaystyle (3n^2 +20n +13)^{\frac{1}{n}} \le (3n^2 + 20n^2 + 13n^2)^{\frac{1}{n}} = (36n^2)^{\frac{1}{n}}$

And then use a similar trick to get an upper bound for the expression, and finally apply the sandwhich theorem.
• Nov 28th 2010, 07:50 AM
elemental
Uh-oh, it's been a while since I've taken calculus. Time to review!
• Nov 28th 2010, 01:24 PM
azarue
thanks for the help
I should practice the sandwich rule more, would it be safe to say it is better to try the sandwich approach with sums ?

could that be used as the inequality for the sandwich ? $\displaystyle \frac{n}{n^2+n}\ge A_n\ge\frac{n^2}{n^2+1}$ for the first one and $\displaystyle 13^{\frac{1}{n}} \le (3n^2 +20n +13)^{\frac{1}{n}} \le (3n^2 + 20n^2 + 13n^2)^{\frac{1}{n}} = (36n^2)^{\frac{1}{n}}$ for the second one.