Moved_Limits

[val-val]

I'm having problems with these questions.

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<sub>I've tried doing them on my own but since I missed the lesson on this I am completely lost. Help!




~~~~~Sorry guys, I put this question in the wrong sections :| Not sure how to delete it though.</sub>

 

[JeffM]

I'm having problems with these questions.

View attachment 2865
_{I've tried doing them on my own but since I missed the lesson on this I am completely lost. Help!}

There is a general rule for rational functions like this, but let's work on specifics so you will see where the general rule comes from

\(\displaystyle f(n) = \dfrac{n^3 - 2}{n^2} = n - \dfrac{2}{n^2} \implies \displaystyle \lim_{n \rightarrow \infty}f(n) = \left(\lim_{n \rightarrow \infty}n\right) - \left(\lim_{n \rightarrow \infty}\dfrac{2}{n^3}\right) = ?\)

Does that help?

 

[val-val]

There is a general rule for rational functions like this, but let's work on specifics so you will see where the general rule comes from

\(\displaystyle f(n) = \dfrac{n^3 - 2}{n^2} = n - \dfrac{2}{n^2} \implies \displaystyle \lim_{n \rightarrow \infty}f(n) = \left(\lim_{n \rightarrow \infty}n\right) - \left(\lim_{n \rightarrow \infty}\dfrac{2}{n^3}\right) = ?\)

Does that help?

I just don't understand where to start. At least I know theres a rule now! How would you start off a problem like this? Do you plug stuff in or is there some assuming to be done?

EDIT: Ohhh, I see what you did. Towards the end, do you just subtract? Would you take n away from 2/n^3 ? So it would end up being 2/n^2 ?
I could be misunderstanding you :/

 

[JeffM]

I just don't understand where to start. At least I know theres a rule now! How would you start off a problem like this? Do you plug stuff in or is there some assuming to be done?

EDIT: Ohhh, I see what you did. Towards the end, do you just subtract? Would you take n away from 2/n^3 ? So it would end up being 2/n^2 ?
I could be misunderstanding you :/

I am afraid you are.

As n gets larger and larger what does \(\displaystyle \dfrac{2}{n}\) become approximately equal to?

If n grows beyond all bounds, what is the limit of n?

There are some fundamental laws of limits that you need to MEMORIZE:

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = c \in \mathbb R \implies \left\{\lim_{x \rightarrow a}f(x) + g(x)\right\} = (b + c) \in \mathbb R.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = c \in \mathbb R \implies \left\{\lim_{x \rightarrow a}f(x) - g(x)\right\} = (b - c) \in \mathbb R.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = c \in \mathbb R \implies \left\{\lim_{x \rightarrow a}f(x) * g(x)\right\} = (b * c) \in \mathbb R.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = c \in \mathbb R\ and\ c \ne 0 \implies \left\{\lim_{x \rightarrow a}\dfrac{f(x)}{g(x)}\right\} = \dfrac{b}{c} \in \mathbb R.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = \infty \implies \left\{\lim_{x \rightarrow a}f(x) + g(x)\right\} = \infty.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = \infty \implies \left\{\lim_{x \rightarrow a}f(x) - g(x)\right\} = - \infty.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = \infty \implies \left\{\lim_{x \rightarrow a}f(x) * g(x)\right\} = \infty.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = \infty \implies \left\{\lim_{x \rightarrow a}\dfrac{f(x)}{g(x)}\right\} = 0.\)

These laws apply whether or not a is a real number or positive or negative infinity.

For many limit problems, these will let you solve them by breaking the limit of a complex function into a combination of limits of simpler functions.

Notice what is not covered by those laws.

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = 0\ and\ \lim_{x \rightarrow a}g(x) = 0.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = \infty\ and\ \lim_{x \rightarrow a}g(x) = \infty.\)

The rule for rational functions is a consequence of these rules

 

[soroban]

Hello, val-val!

With problems like these, divide numerator and denominator
by the highest power of the variable in the denominator.

\(\displaystyle \displaystyle 1.\;\lim_{n\to\infty}\frac{n^3-2}{n^2}\)

Divide numerator and denominator by \(\displaystyle n^2\)

. . \(\displaystyle \displaystyle \frac{\frac{n^3}{n^2} - \frac{2}{n^2}}{\frac{n^2}{n^2}} \;=\;\frac{n - \frac{2}{n^2}}{1}\)

Therefore: .\(\displaystyle \displaystyle \lim_{n\to\infty}\left(n - \tfrac{2}{n^2}\right) \;=\;\infty - 0 \;=\;\infty\)

\(\displaystyle \displaystyle 2.\;\lim_{n\to\infty}\frac{8n^2+5n+2}{5+2n}\)

Divide numerator and denominator by \(\displaystyle n.\)

. . \(\displaystyle \displaystyle \frac{\frac{8n^2}{n} + \frac{5n}{n} + \frac{2}{n}}{\frac{5}{n} + \frac{2n}{n}} \;=\;\frac{8n + 5 + \frac{2}{n}}{\frac{5}{n} + 2}\)

Therefore: .\(\displaystyle \displaystyle \lim_{n\to\infty}\frac{8n + 5 + \frac{2}{n}}{\frac{5}{n} + 2} \;=\;\frac{\infty + 5 + 0}{0 + 2} \;=\;\infty\)