2.1.2 Limits at infinity

It might not look like a traditional polynomial.

There are many answers: any constant polynomial has this property. One answer is $f(x)=1$.

Any polynomial of degree one or higher will go to $\infty$ or $-\infty$ as $x$ goes to $\infty$. So, we need a polynomial of degree 0–that is, $f(x)$ is a constant. One possible answer is $f(x)=1$.

The degree of the polynomial matters.

There are many answers: any odd-degree polynomial has this property. One answer is $f(x)=x$.

This will be the case for any polynomial of odd degree. For instance, $f(x)=x$.

Many answers are possible: also $f(x)=x^{15}-32x^2+9$ satisfies $\displaystyle\lim_{x \rightarrow \infty} f(x) = \infty$ and $\displaystyle\lim_{x \rightarrow -\infty} f(x)= -\infty$.

What does a negative exponent do?

0

$\displaystyle\lim_{x \rightarrow \infty} 2^{-x}= \displaystyle\lim_{x \rightarrow \infty} \frac{1}{2^x}=0$

You can think about the behaviour of this function by remembering how you first learned to describe exponentiation.

$\infty$

As $x$ gets larger and larger, $2^x$ grows without bound. (For integer values of $x$, you can imagine multiplying $2$ by itself more and more times.) So, $\displaystyle\lim_{x \rightarrow \infty}2^x=\infty$.

The exponent will be a negative number.

$0$

Write $X=-x$. As $x$ becomes more and more negative, $X$ becomes more and more positive. From Question 4, we know that $2^X$ grows without bound as $X$ gets larger and larger. Since $2^x= 2^{-(-x)}= 2^{-X} =\frac{1}{2^X}$, as we let $x$ become a huge negative number, we are in effect dividing by a huge positive number; hence $\lim\limits_{x\rightarrow-\infty} 2^x = 0$.

A more formulaic way to describe the above is this: $\lim\limits_{x\rightarrow-\infty} 2^x = \lim\limits_{X\rightarrow\infty} 2^{-X} = \lim\limits_{X\rightarrow\infty} \frac{1}{2^X} =0$.

What single number is the function approaching?

DNE

There is no single number that $\cos x$ approaches as $x$ becomes more and more strongly negative: as $x$ grows in the negative direction, the function oscillates between $-1$ and $+1$, never settling close to one particular number. So, this limit does not exist.

The highest-order term dominates when $x$ is large.

$-\infty$

The highest-order term in this polynomial is $-3x^5$, so this dominates the function's behaviour as $x$ goes to infinity. Therefore, the limit is $-\infty$.

Another way to see this:

because

Factor the highest power of $x$ out of both the numerator and the denominator. You can factor through square roots (carefully).

$\sqrt{3}$

Solution 1
For large values of $x$, recall that higher powers dominate. So:

Therefore, $\lim\limits_{x \rightarrow\infty} \dfrac{\sqrt{3x^8+7x^4}+10}{x^4-2x^2+1}=\sqrt3$.

Solution 2
Our other standard trick is to factor out the highest power of $x$ in the denominator: $x^4$. We just have to be a little careful with the square root. Since we are taking the limit as $x$ goes to positive infinity, we have positive $x$-values, so $\sqrt{x^2}=x$ and $\sqrt{x^8}=x^4$.

Multiply and divide by the conjugate, $\sqrt{x^2+5x}+\sqrt{x^2-x}$.

3

First, a note about asyptotics. It's tempting (but wrong!) to write the following:

Recall Example 1.1.5 in the text for a concrete example: subtracting two expressions that are asymptotically the same doesn't necessarily give you an expression tending to 0.

More generally, here's the source of a common mistake. By Theorem 2.1.15, if $\lim\limits_{x \to a} f(x)=F$ and $\lim\limits_{x \to a}g(x)=G$, and $F$ and $G$ are both real numbers, then you can conclude that $\lim\limits_{x \to a}[f(x)-g(x)]=F-G$. However, if $F$ and $G$ are not real numbers (for example, if they're both infinite), the theorem does not apply.

Now that we know how not to solve this question, let's actually solve it!

We have two terms, each getting extremely large. It's unclear at first what happens when we subtract them. To get this equation into another form, we multiply and divide by the conjugate, $\sqrt{x^2+5x}+\sqrt{x^2-x}$.

Now we divide the numerator and denominator by $x$. In the case of the denominator, since $x>0$, $x=\sqrt{x^2}$.

Divide both the numerator and the denominator by the highest power of $x$ that is in the denominator.
Remember that $\sqrt{\ }$ is defined to be the positive square root. Consequently, if $x<0$, then $\sqrt{x^2}$, which is positive, is not the same as $x$, which is negative.

$-\frac{3}{4}$

Note that for large negative $x$, the first term in the denominator $\sqrt{4x^2+x}\approx\sqrt{4x^2}=|2x|=-2x$ not $+2x$. A good way to avoid incorrectly computing $\sqrt{x^2}$ when $x$ is negative is to define $y=-x$ and express everything in terms of $y$. That's what we'll do.

Factor out the highest power of the denominator.

$-\dfrac{1}{2}$

Solution 1
For $x$ far from 0, the numerator is dominated by $-x^2$ and the denominator by $2x^2$, so

Solution 2
The highest power of $x$ in the denominator is $x^2$, so we divide the numerator and denominator by $x^2$:

The conjugate of $(\sqrt{x^2+x}-x)$ is $(\sqrt{x^2+x}+x)$.
Multiply by $1=\dfrac{\sqrt{x^2+x}+x} {\sqrt{x^2+x}+x}$ to coax your function into a fraction.

$\frac{1}{2}$

Remark: see Example 1.1.5 in the text for a concrete example of why subtracting two expressions that are asymptotically the same doesn't necessarily give you an expression tending to 0.

Divide both the numerator and the denominator by the highest power of $x$ that is in the denominator.

$\frac{5}{3}$

Solution 1
For large $x$, the numerator behaves asymptotically like $5x^2$, and the denominator like $3x^2$, so $\displaystyle \lim_{x\to +\infty} \frac{5x^2-3x+1}{3x^2 +x+7}=\lim_{x \to \infty}\frac{5x^2}{3x^2}=\frac53.$

Solution 2
We have, after dividing both numerator and denominator by $x^2$ (which is the highest power of the denominator) that

Since $1/x\to 0$ and also $1/x^2\to 0$ as $x\to +\infty$, we conclude that

Divide both the numerator and the denominator by the highest power of $x$ that is in the denominator.

0

Solution 1
For large values of $x$:

so the limit is 0.

Solution 2
We have, after dividing both numerator and denominator by $x$ (which is the highest power of the denominator) that

Since $1/x\to 0$ and also $1/x^2\to 0$ as $x\to +\infty$, we conclude that

Divide both the numerator and the denominator by the highest power of $x$ that is in the denominator.

$\frac{4}{7}$

Solution 1
For large $x$, the numerator is dominated by $4x^3$, and the denominator by $7x^3$. So:

Solution 2
The dominant terms in the numerator and denominator have order $x^3$. Taking out that common factor we get

Since $1/x^a\to 0$ as $x\to +\infty$ (for $a>0$), we conclude that

Divide both the numerator and the denominator by $x$ (which is the largest power of $x$ in the denominator). In the numerator, move the resulting factor of $1/x$ inside the two roots. Be careful about the signs when you do so. Even and odd roots behave differently– see Question 10.

1

Use asymptotics, or divide both the numerator and the denominator by the highest power of $x$ that is in the denominator.

0

Solution 1
For large values of $x$, the numerator behaves asymptotically like $5x^2$, while the denominator behaves like $3x^3$. Larger exponents dominate when $x$ is large, so the denominator grows much, much faster than the numerator, so the limit is 0.

Solution 2
We have, after dividing both numerator and denominator by $x^3$ (which is the highest power of the denominator) that:

Divide both the numerator and the denominator by the highest power of $x$ that is in the denominator. It is not always true that $\sqrt{x^2}=x$.

$-1$

Since we only consider negative values of $x$, $\sqrt{x^2}=|x|=-x$.

Simplify.

$1$

Since we only consider positive values of $x$, $\sqrt{x^2}=|x|=x$.

What is a simpler version of $|x|$ when you know $x<0$?

$-1$

When $x<0$, $|x|=-x$ and so $\ds\lim_{x\to\infty}\sin\left( \frac{\pi}{2} \cdot \frac{|x|}{x}\right) + \frac{1}{x} = \sin(-\pi/2) = -1$.

$-\frac{3}{2}$

We divide both the numerator and the denominator by the highest power of $x$ in the denominator, which is $x$. Since $x<0$, we have $\sqrt{x^2}=|x|=-x$, so that

Since $1/x\to 0$ and also $1/x^2\to 0$ as $x\to -\infty$, we conclude that

Remark: see Example 1.1.5 in the text for a concrete example of why subtracting two expressions that are asymptotically the same doesn't necessarily give you an expression tending to 0. That's why we can't assume the denominator approaches 0.

Divide both the numerator and the denominator by the highest power of $x$ that is in the denominator. When is $\sqrt{x}=x$, and when is $\sqrt{x}=-x$?

$-\frac{5}{3}$

Solution 1
We divide both the numerator and the denominator by the highest power of $x$ in the denominator, which is $x$. Since $x<0$, we have $\sqrt{x^2}=|x|=-x$, so that

Since $1/x\to 0$ and also $1/x^2\to 0$ as $x\to -\infty$, we conclude that

Solution 2
For $x$ far from 0, we have:

So,

Divide both the numerator and the denominator by the highest power of $x$ that is in the denominator. Pay careful attention to signs.

$-\infty$

Solution 1
For large $x$, the numerator is dominated by $3x^7$, while the denominator is dominated by $4x^2$. So:

Solution 2

Multiply and divide the expression by its conjugate, $\big(\sqrt{n^2+5n}+n\big)$.

$\dfrac{5}{2}$

We multiply and divide the expression by its conjugate, $\big(\sqrt{n^2+5n}+n\big)$.

Since $n>0$, we can simplify $\sqrt{n^2}=n$.

Remark: see Example 1.1.5 in the text for a concrete example of why subtracting two expressions that are asymptotically the same doesn't necessarily give you an expression tending to 0.

Consider what happens to the function as $a$ becomes very, very small. You shouldn't need to do much calculation.

$\ds\lim_{a \to 0^+}\dfrac{a^2-\frac{1}{a}}{a-1}=\infty$

Since $x=3$ is not in the domain of the function, we need to be a little creative. Try simplifying the function.

$\ds\lim_{x \to 3}\dfrac{2x+8}{\frac{1}{x-3}+\frac{1}{x^2-9}}=0$

Since $x=3$ is not in the domain of the function, we simplify, hoping we can cancel a problematic term.