2.1.1 Calculating limits with limit laws

(a) and (d)

Zeroes cause a problem when they show up in the denominator, so we can only compute (a) and (d). (Both these limits are zero.) Be careful: there is no such rule as “zero divided by zero is one," or “zero divided by zero is zero."

Try to make two functions with factors that will cancel.

There are many possible answers; one is $f(x)=10(x-3)$, $g(x)=x-3$.

The statement $\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=10$ tells us that, as $x$ gets very close to 3, $f(x)$ is 10 times as large as $g(x)$. We notice that if $f(x)=10g(x)$, then $\dfrac{f(x)}{g(x)}=10$, so $\displaystyle\lim_{x \rightarrow} \dfrac{f(x)}{g(x)}=10$ wherever $f$ and $g$ exist. So it's enough to find a function $g(x)$ that has limit 0 at 3. Such a function is (for example) $g(x)=x-3$. So, we take $f(x)=10(x-3)$ and $g(x)=x-3$. It is easy now to check that $\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$ and $\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=\ds\lim_{x \to 3}\frac{10(x-3)}{x-3}=\ds\lim_{x \to 3}10=10$.

Try to make $g(x)$ cancel out.

There are many possible answers; one is $f(x)=(x-3)^2$ and $g(x)=x-3$. Another is $f(x)=0$ and $g(x)=x-3$.

There are many more possible answers.

There are many possible answers; one is $f(x)=x-3$, $g(x)=(x-3)^3$.

One way to start this problem is to remember $\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x^2}=\infty$. (Using $\dfrac{1}{x^2}$ as opposed to $\dfrac{1}{x}$ is important, since $\displaystyle\lim_{x \rightarrow 0}\dfrac{1}{x}$ does not exist.) Then by “shifting" by three, we find $\displaystyle \lim_{x \rightarrow 3} \dfrac{1}{(x-3)^2}=\infty$. So it is enough to arrange that $\dfrac{f(x)}{g(x)}=\dfrac{1}{(x-3)^2}$. We can achieve this with $f(x)=x-3$ and $g(x)=(x-3)^3$, and maintain $\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$. Again, this is one of many possible solutions.

See Questions 2, 3, and 4.

Any real number; positive infinity; negative infinity; does not exist.

Any real number; positive infinity; negative infinity; does not exist.
This is an important thing to remember: often, people see limits that look like $\dfrac{0}{0}$ and think that the limit must be 1, or 0, or infinite. In fact, this limit could be anything–it depends on the relationship between $f$ and $g$.

Questions 2 and 3 show us examples where the limit is 10 and 0; they can easily be modified to make the limit any real number.

Question 4 show us an example where the limit is $\infty$; it can easily be modified to make the limit $-\infty$ or DNE.

Find the limit of the numerator and denominator separately.

0

Since we're not trying to divide by 0, or multiply by infinity:$\displaystyle\lim_{t \rightarrow 10} \dfrac{2(t-10)^2}{t}

\dfrac{2\cdot0}{10}=0$

Break it up into smaller pieces, evaluate the limits of the pieces.

6

Since we're not doing anything dodgy like putting 0 in the denominator,
$\displaystyle\lim_{y \rightarrow 0} \dfrac{(y+1)(y+2)(y+3)}{\cos y} =\dfrac{(0+1)(0+2)(0+3)}{\cos 0}=\dfrac{6}{1}=6$.

First find the limit of the “inside" function, $\dfrac{4x-2}{x+2}$.

16

Since the limits of the numerator and denominator exist, and since the limit of the denominator is nonzero: $\displaystyle\lim_{x \rightarrow 3} \left(\dfrac{4x-2}{x+2}\right)^4 = \left(\dfrac{4(3)-2}{3+2}\right)^4=16$

Is $\cos(-3)$ zero?

$4/\cos(3)$

Expand, then simplify.

$2$

If try naively then we get $0/0$, so we expand and then simplify:

Hence the limit is $\ds \lim_{h \to 0} \left(\frac{h}{2}+2\right) = 2$.

$-7/2$

3

$-\frac{3}{2}$

Try the simplest method first.

$\log(2)-1$

We simply plug in $x=1$: $\lim\limits_{x\rightarrow 1}\left[\dfrac{\log(1+x)-x}{x^2}\right]=\log(2)-1$.

Factor the denominator.

$\frac{1}{4}$

If we try naively then we get $0/0$, so we simplify first:

Hence the limit is $\displaystyle \lim_{x\rightarrow2} \frac{1}{x+2} = 1/4$.

Factor the numerator and the denominator.

$\dfrac{1}{2}$

If we try to plug in $x=4$, we find the denominator is zero. So to get a better idea of what's happening, we factor the numerator and denominator:

Factor the numerator.

$5$

If we try to plug in $x=2$, we find the denominator is zero. So to get a better idea of what's happening, we factor the numerator:

Simplify first by factoring the numerator.

$-6$

If we try naively then we get $0/0$, so we simplify first:

Hence the limit is $\ds \lim_{x\to-3} (x-3) = -6$.

The function is a polynomial.

$-14$

To calculate the limit of a polynomial, we simply evaluate the polynomial:
$\displaystyle\lim_{t \rightarrow 2} \frac{1}{2}t^4-3t^3+t

\frac{1}{2}\cdot2^4-3\cdot 2^3+2 = -14$

Multiply both the numerator and the denominator by the conjugate of the numerator, $\sqrt{x^2+8}+3$.

$-\frac{1}{3}$

Multiply both the numerator and the denominator by the conjugate of the numerator, $\sqrt{x+2}+\sqrt{4-x}$.

$\frac{1}{\sqrt{3}}$

If we try to do the limit naively we get $0/0$. Hence we must simplify.

So the limit is

Multiply both the numerator and the denominator by the conjugate of the numerator, $\sqrt{x-2}+\sqrt{4-x}$.

1

If we try to do the limit naively we get $0/0$. Hence we must simplify.

Consider the factors $x^2$ and $\cos\left(\frac{3}{x}\right)$ separately.

0

First, let's think of some general principles.

For any nonzero value of $x$ (whether or not it's close to 0), $\left|\cos\left(\frac3x\right) \right|\le 1$. So, its magnitude never gets very large. Since it's multiplied by something going to 0, the entire function will go to 0.

We can also see this by graphing the function. Note that $\cos\left(\frac3x\right)$ keeps cycling from 1, to 0, to -1, back to 0, etc, as $x$ approaches 0.

So, we imagine the function $x^2\cos\left(\frac3x\right)$ wiggling back and forth between $x^2$ and $-x^2$:

For $x$ very close to 0, then, also $-x^2\cos\left(\frac3x\right)$ is very close to 0. That is, $\displaystyle\lim_{x \rightarrow 0}-x^2\cos\left(\frac{3}{x}\right)=0$.

Compare to the previous question.

0

For any (nonzero) value of $x$, $0 \le 1\sin^2\left(\dfrac{1}{x}\right)\le 1$. So when we multiply it by a number, that number either stays the same or gets closer to 0.

In particular, when we multiply $x$ by $\sin^2\left(\dfrac{1}{x}\right)$, the result is either $x$ itself, or something even closer to 0 than $x$ was originally. Since $x$ is approaching 0, $x\sin^2\left(\dfrac{1}{x}\right)$ is approaching 0 as well. That is, $\lim\limits_{x\rightarrow 0}x\sin^2\left(\dfrac{1}{x}\right)=0$.

Another way to see this is by graphing. The factor $\sin^2\left(\dfrac{1}{x}\right)\le 1$ cycles between 0 and 1, so the function $x\sin^2\left(\dfrac{1}{x}\right)$ cycles between 0 and $x$:

Again, we see $\lim\limits_{x\rightarrow 0}x\sin^2\left(\dfrac{1}{x}\right)=0$.

Factor the numerator.

5

When we plug $w=5$ in to the numerator and denominator, we find that each becomes zero. Since we can't divide by zero, we have to dig a little deeper. When a polynomial has a root, that also means it has a factor: we can factor $(w-5)$ out of the top. That lets us cancel:

Note that the function $\dfrac{2w^2-50}{(w-5)(w-1)}$ is NOT defined at $w=5$, while the function $\dfrac{2(w+5)}{(w-1)}$ IS defined at $w=5$; so strictly speaking, these two functions are not equal. However, for every value of $w$ that is not 5, the functions are the same, so their limits are equal. Furthermore, the limit of the second function is quite easy to calculate, since we've eliminated the zero in the denominator: $\displaystyle\lim_{w \rightarrow 5} \dfrac{2(w+5)}{(w-1)} =\dfrac{2(5+5)}{5-1}=5.$

So $\displaystyle\lim_{w \rightarrow 5} \dfrac{2w^2-50}{(w-5)(w-1)}=\displaystyle\lim_{w \rightarrow 5} \dfrac{2(w+5)}{(w-1)}=5$.

Factor the denominator; pay attention to signs.

$-\infty$

When we plug in $r=-5$ to the denominator, we find that it becomes 0, so we need to dig deeper. The numerator is not zero, so cancelling is out. Notice that the denominator is factorable: $r^2+10r+25 = (r+5)^2$. As $r$ approaches $-5$ from either side, the denominator gets very close to zero, but stays positive. The numerator gets very close to $-5$. So, as $r$ gets closer to $-5$, we have something close to $-5$ divided by a very small, positive number. Since the denominator is small, the fraction will have a large magnitude; since the numerator is negative and the denominator is positive, the fraction will be negative. So, $\displaystyle\lim_{r \rightarrow -5} \dfrac{r}{r^2+10r+25}=-\infty$

First find the limit of the “inside" function.

$\sqrt{\dfrac{2}{3}}$

First, we find $\displaystyle\lim_{x \rightarrow -1}\dfrac{x^3+x^2+x+1}{3x+3}$. When we plug in $x=-1$ to the top and the bottom, both become zero. In a polynomial, where there is a root, there is a factor, so this tells us we can factor out $(x+1)$ from both the top and the bottom. It's pretty easy to see how to do this in the bottom. For the top, if you're having a hard time, one factoring method (of many) to try is long division of polynomials; another is to factor out $(x+1)$ from the first two terms and the last two terms. (Detailed examples of long division are given in Appendix A.16 and Examples 1.10.2 and 1.10.3 of CLP–2.)

One thing to note here is that the function $\dfrac{x^3+x^2+x+1}{3x+3}$ is not defined at $x=-1$ (because we can't divide by zero). So we replaced it with the function $\dfrac{x^2+1}{3}$, which IS defined at $x=-1$. These functions only differ at $x=-1$; they are the same at every other point. That is why we can use the second function to find the limit of the first function.

Now we're ready to find the actual limit asked in the problem:

Factor; pay attention to signs.

DNE

When we plug $x=0$ into the denominator, we get 0, which means we need to look harder. The numerator is not zero, so we won't be able to cancel our problems away. Let's factor to make things clearer.

As $x$ gets close to 0, the numerator is close to 1; the term $(3x^2-5)$ is negative; and the sign of $x^3$ depends on the direction we're approaching 0 from. Since we're dividing a numerator that is very close to 1 by something that's getting very close to 0, the magnitude of the fraction is getting bigger and bigger without bound. Since the sign of the fraction flips depending on whether we are using numbers slightly bigger than 0, or slightly smaller than 0, that means the one-sided limits are $\infty$ and $-\infty$, respectively. (In particular, $\displaystyle\lim_{x \rightarrow 0^-}\dfrac{x^2+2x+1}{3x^5-5x^3}=\infty$ and $\displaystyle\lim_{x \rightarrow 0^+}\dfrac{x^2+2x+1}{3x^5-5x^3}=-\infty$.) Since the one-sided limits don't agree, the limit does not exist.

Look for perfect squares

$\infty$

As usual, we first try plugging in $t=7$, but the denominator is 0, so we need to think harder. The top and bottom are both squares, so let's go ahead and factor: $\dfrac{t^2x^2+2tx+1}{t^2-14t+49}= \dfrac{(tx+1)^2}{(t-7)^2}$. Since $x$ is positive, the numerator is nonzero. Also, the numerator is positive near $t=7$. So, we have something positive and nonzero on the top, and we divide it by the bottom, which is positive and getting closer and closer to zero. The quotient is always positive near $t=7$, and it is growing in magnitude without bound, so $\displaystyle\lim_{t \rightarrow 7} \dfrac{t^2x^2+2tx+1}{t^2-14t+49}=\infty$.

Remark: there is an important reason we specified that $x$ must be a positive constant. Suppose $x$ were $-\frac{1}{7}$ (which is negative and so was not allowed in the question posed). In this case, we would have

Think about what effect changing $d$ has on the function $x^5-32x+15$.

$x^5-32x+15$

The function whose limit we are taking does not depend on $d$. Since $x$ is a constant, $x^5-32x+15$ is also a constant–it's just some number, that doesn't change, regardless of what $d$ does. So $\displaystyle\lim_{d \rightarrow 0} x^5-32x+15=x^5-32x+15$.

There's an easy way.

$0$

There's a lot going on inside that sine function... and we don't have to care about any of it. No matter what horrible thing we put inside a sine function, the sine function will spit out a number between $-1$ and $1$. So that means the entire function is somewhere between $(x-1)^2$ and $-(x-1)^2$. Since $(x-1)^2$ is approaching 0, the entire function is approaching 0.

That is, $\displaystyle\lim_{x \rightarrow 1} (x-1)^2\sin\left[\left(\dfrac{x^2-3x+2}{x^2-2x+1}\right)^2+15\right]=0$.

What can you do to safely ignore the sine function?

$0$

Since $-1 \leq \sin x \leq 1$ for all values of $x$, when we multiply a number by this function, it causes the magnitude (absolute value) of that number to either be the same, or closer to 0.

Since $\lim\limits_{x \to 0} x^{1/101}$ is already 0, the limit doesn't change when we multiply it by the sine part.

Factor

2

If you're looking at the hints for this one, it's probably easier than you think.

0

When we plug in $x=5$ to the top and the bottom, both limits exist, and the bottom is nonzero. So $\displaystyle\lim_{x \rightarrow 5} \dfrac{(x-5)^2}{x+5}= \dfrac{0}{10}=0$.

You'll want to simplify this, since $t=\frac{1}{2}$ is not in the domain of the function. One way to start your simplification is to add the fractions in the numerator by finding a common denominator.

$-\dfrac{32}{9}$

Since we can't plug in $t=\frac{1}{2}$, we'll simplify. One way to start is to add the fractions in the numerator. We'll need a common demoninator, such as $3t^2(t^2-1)$.

Since we cancelled out the term that was causing the numerator and denominator to be zero when $t=\frac{1}{2}$, now $t=\frac{1}{2}$ is in the domain of our function, so we simply plug it in: