Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 3

Limits

Infinite Limits: Vertical Asymptotes

In this section, we discuss evaluating limits $\underset{x\to a}{\lim }f\left(x\right)\text{,}$ where the function $f$ is neither continuous at $a$ nor capable of being simplified into another function (as suggested in the previous section). For such limits, we must analyse the function.

We start with the problem of sketching the graph of the reciprocal function:

${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}}$

if the graph of $g\left(x\right)$ is known.

Study the following statements about the reciprocal function. You must understand and agree with them, not just believe or accept them.

• If $g\left(a\right)=0$, then

  ${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}}$

  is not defined at $a\text{.}$

• If $g\left(x\right)>0$ on an interval $J$, then

  ${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}>0}$

  on the same interval $J\text{.}$

• If $g\left(x\right)<0$ on an interval $J$, then

  ${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}<0}$

  on the same interval $J\text{.}$

• If $g\left(a\right)$ is positive and near $0$, then

  ${\displaystyle f\left(x\right)=\frac{1}{g\left(a\right)}}$

  is positive and very large (far from $0$).

• If $g\left(a\right)$ is negative and near $0$, then

  ${\displaystyle f\left(x\right)=\frac{1}{g\left(a\right)}}$

  is negative and very small (far from $0$).

What is ${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}}$ like if $g\left(x\right)$ is positive and very large (far from $0$)?

What about if $g\left(x\right)$ is negative and very small (far from $0$)?

Example 3.39. The graph of the identity function $g\left(x\right)=x$ is shown in Figure 3.18, below.

Figure 3.18. Function $g\left(x\right)=x$

From the statements above, we know the following facts about the identity function’s reciprocal function, $f\left(x\right)=1∕x$:

• $g\left(0\right)=0$; hence, ${\displaystyle f\left(x\right)=\frac{1}{x}}$ is not defined at $0\text{.}$
• $g\left(x\right)>0$ on $\left(0\text{,}\infty \right)\text{;}$ hence, ${\displaystyle f\left(x\right)=\frac{1}{x}>0}$ for $x>0\text{.}$
• $g\left(x\right)<0$ on $\left(-\infty \text{,}0\right)$; hence, ${\displaystyle f\left(x\right)=\frac{1}{x}<0}$ for $x<0\text{.}$
• $g\left(x\right)>0$ and very small for $x\to 0^{+}$; hence, ${\displaystyle f\left(x\right)=\frac{1}{x}>0}$ and very large (far from $0$) for $x\to 0^{+}\text{.}$
• $g\left(x\right)<0$ and very large for $x\to 0^{-}$; hence, ${\displaystyle f\left(x\right)=\frac{1}{x}<0}$ and very small (far from $0$) for $x\to 0^{-}\text{.}$
• ${\displaystyle f\left(x\right)=\frac{1}{x}>0}$ and near to $0$, if $x$ is very large and positive (far from $0$). [Why?]
• ${\displaystyle f\left(x\right)=\frac{1}{x}<0}$ and near to $0$, if $x$ is very small and negative (far from $0$). [Why?]

Using this information, we see that the graph of ${\displaystyle f\left(x\right)=\frac{1}{x}}$ is as shown in Figure 3.19, below.

Figure 3.19. Function ${\displaystyle f\left(x\right)=\frac{1}{x}}$

Example 3.40. If $g\left(x\right)=x^2-1$, then the graph of $g$ is as shown in Figure 3.20, below, and its reciprocal is the function ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}\text{,}}$ shown in Figure 3.21. Observe why.

Figure 3.20. Function $g\left(x\right)=x^2-1$

Applying the statements above to the graph in Figure 3.20, we find the following facts about its reciprocal:

• ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}}$ is not defined at $1$ or $-1\text{.}$
• ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}>0}$ on $\left(-\infty \text{,}-1\right)$ and $\left(1\text{,}\infty \right)\text{.}$
• ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}<0}$ on $\left(-1\text{,}1\right)\text{.}$
• ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}>0}$ and far from $0$ for $x\to 1^{+}$ and for $x\to -1^{-}\text{.}$
• ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}<0}$ and far from $0$ for $x\to 1^{-}$ and for $x\to -1^{+}\text{.}$
• ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}>0}$ and very small (close to $0$) for very large positive $x$, and very small negative $x\text{.}$

Using this information, we determine that the graph ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}}$ is as shown in Figure 3.21, below.

Figure 3.21. Function ${\displaystyle f\left(x\right)=\frac{1}{x^2-1}}$

Example 3.41. In Figure 3.22, below, we show the graph of a function $g\left(x\right)$, and in Figure 3.23 we show the graph of its reciprocal ${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}}\text{.}$ Examine these graphs. Do you agree with them?

Figure 3.22. Function $g\left(x\right)$

Figure 3.23. Function ${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}}$

Example 3.42. In Figure 3.24, below, we show the graph of the function $g\left(x\right)=\left|x\right|$, and in Figure 3.25, the graph of its reciprocal. Examine these graphs. Do you agree with them?

Figure 3.24. Function $g\left(x\right)=\left|x\right|$

Figure 3.25. Function ${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}}$; that is, ${\displaystyle f\left(x\right)=\frac{1}{\left|x\right|}}$

From the graphs of Examples 3.39 to 3.42, we derive the following theorem.

Theorem 3.17.

1. If $\underset{x\to a^{+}}{\lim }g\left(x\right)=0$ and $g\left(x\right)>0$ for $x\to a^{+}$, then ${\displaystyle \underset{x\to a^{+}}{\lim }\frac{1}{g\left(x\right)}=\infty \text{.}}$
2. If $\underset{x\to a^{-}}{\lim }g\left(x\right)=0$ and $g\left(x\right)>0$ for $x\to a^{-}$, then ${\displaystyle \underset{x\to a^{-}}{\lim }\frac{1}{g\left(x\right)}=\infty \text{.}}$
3. If $\underset{x\to a^{+}}{\lim }g\left(x\right)=0$ and $g\left(x\right)<0$ for $x\to a^{+}$, then ${\displaystyle \underset{x\to a^{+}}{\lim }\frac{1}{g\left(x\right)}=-\infty \text{.}}$
4. If $\underset{x\to a^{-}}{\lim }g\left(x\right)=0$ and $g\left(x\right)<0$ for $x\to a^{-}$, then ${\displaystyle \underset{x\to a^{-}}{\lim }\frac{1}{g\left(x\right)}=-\infty \text{.}}$

   Therefore,

5. ${\displaystyle \underset{x\to a}{\lim }\frac{1}{g\left(x\right)}=\infty }$, if $\underset{x\to a}{\lim }g\left(x\right)=0$ and $g\left(x\right)>0$ for $x\to a\text{.}$
6. ${\displaystyle \underset{x\to a}{\lim }\frac{1}{g\left(x\right)}=-\infty }$, if $\underset{x\to a}{\lim }g\left(x\right)=0$ and $g\left(x\right)<0$ for $x\to a\text{.}$

Example 3.43. The function ${\displaystyle f\left(x\right)=\frac{1}{\sqrt{9+x}-3}}$ in the limit

${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{1}{\sqrt{9+x}-3}}$

is the reciprocal of the function $g\left(x\right)=\sqrt{9+x}-3\text{.}$

However, $\underset{x\to 0^{+}}{\lim }\sqrt{9+x}-3=0$ and $\sqrt{9+x}-3>0$, for $x\to 0^{+}$; hence, by Theorem 3.18(a), we conclude that

${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{1}{\sqrt{9+x}-3}=\infty \text{.}}$

Note: To see that $\color{#384877}{\sqrt{9 + x} -3 \gt 0}$ for $\color{#384877}{x \to 0^+}$, take an $\color{#384877}{x}$ positive and close to $\color{#384877}{0}$, say $\color{#384877}{x = 0.1}$, and check that $\color{#384877}{\sqrt{9 + 0.1} - 3 \gt 0}$.

Example 3.44. By Theorem 3.17,

${\displaystyle \underset{x\to 0^{-}}{\lim }\frac{1}{\sqrt{x+9}-3}=-\infty \text{,}}$

because $\underset{x\to 0^{-}}{\lim }\sqrt{x+9}-3=0\text{,}$ and $\sqrt{x+9}-3<0$ for $x\to 0^{-}\text{.}$

From these two examples, we conclude that the limit

${\displaystyle \underset{x\to 0}{\lim }\frac{1}{\sqrt{x+9}-3}}$

does not exist.

Example 3.45.

${\displaystyle \underset{x\to 1^{-}}{\lim }\frac{1}{x^2+x-2}=-\infty }$

because $\underset{x\to 1^{-}}{\lim }{x}^2+x-2=0$ and $x^2+x-2<0$ for $x\to 1^{-}\text{.}$

Example 3.46.

${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{1}{\sin x+\tan x}=\infty }$

because $\underset{x\to 0^{+}}{\lim }\sin x+\tan x=0$, and $\sin x+\tan x>0$ for $x\to 0^{+}\text{.}$

Now, let us analyse a limit of the form $\underset{x\to a}{\lim }h\left(x\right)g\left(x\right)$, where

$\underset{x\to a}{\lim }g\left(x\right)=L\left(L\ne 0\right)\text{,}\text{and}\underset{x\to a}{\lim }h\left(x\right)=\pm \infty \text{.}$

We have several cases to consider.

1. $\underset{x\to a}{\lim }g\left(x\right)=L>0$ and $\underset{x\to a}{\lim }h\left(x\right)=\infty$

   For $x\to a$, we have $g\left(x\right)\to L$ and $h\left(x\right)\to \infty \text{.}$ So, for $x$ close to $a$, $g\left(x\right)$ is close to $L>0$, and $h\left(x\right)$ is very large and positive; hence, the product $g\left(x\right)h\left(x\right)$ is positive and very large, and we have

   $\underset{x\to a}{\lim }h\left(x\right)g\left(x\right)=\infty \text{.}$

2. $\underset{x\to a}{\lim }g\left(x\right)=L<0$ and $\underset{x\to a}{\lim }h\left(x\right)=-\infty$

   For $x\to a$, we have $g\left(x\right)\to L$ and $h\left(x\right)\to -\infty \text{.}$ So, for $x$ close to $a$, $g\left(x\right)$ is close to $L<0$, and $h\left(x\right)$ is very small and negative; hence, the product $g\left(x\right)h\left(x\right)$ is positive and very large, and we have

   $\underset{x\to a}{\lim }h\left(x\right)g\left(x\right)=\infty \text{.}$

3. $\underset{x\to a}{\lim }g\left(x\right)=L>0$ and $\underset{x\to a}{\lim }h\left(x\right)=-\infty$

   For $x\to a$, we have $g\left(x\right)\to L$ and $h\left(x\right)\to -\infty \text{.}$ So, for $x$ close to $a$, $g\left(x\right)$ is close to $L>0$, and $h\left(x\right)$ is very small and negative; hence, the product $g\left(x\right)h\left(x\right)$ is negative and very small, and we have

   $\underset{x\to a}{\lim }h\left(x\right)g\left(x\right)=-\infty$

4. $\underset{x\to a}{\lim }g\left(x\right)=L<0$, and $\underset{x\to a}{\lim }h\left(x\right)=\infty$

   For $x\to a$ we have $g\left(x\right)\to L$ and $h\left(x\right)\to \infty \text{.}$ So, for $x$ close to $a$, $g\left(x\right)$ is close to $L<0$ and $h\left(x\right)$ is very large and positive; hence, the product $g\left(x\right)h\left(x\right)$ is negative and very small, and we have

   $\underset{x\to a}{\lim }h\left(x\right)g\left(x\right)=-\infty$

Therefore, we have Theorem 3.18.

Theorem 3.18.

1. The limit $\underset{x\to a}{\lim }h\left(x\right)g\left(x\right)=\infty$ if either

   1. $\begin{array}{llll}\underset{x\to a}{\lim }g\left(x\right)=L>0\hfill & \text{and}\hfill & \underset{x\to a}{\lim }h(x)=\infty \hfill & \text{or}\hfill \end{array}$

   2. $\begin{array}{lll}\underset{x\to a}{\lim }g\left(x\right)=L<0\hfill & \text{and}\hfill & \underset{x\to a}{\lim }h(x)=-\infty \hfill \end{array}$

2. The limit $\underset{x\to a}{\lim }h\left(x\right)g\left(x\right)=-\infty$ if either

   1. $\begin{array}{llll}\underset{x\to a}{\lim }g\left(x\right)=L<0\hfill & \text{and}\hfill & \underset{x\to a}{\lim }h(x)=\infty \hfill & \text{or}\hfill \end{array}$

   2. $\begin{array}{lll}\underset{x\to a}{\lim }g\left(x\right)=L>0\hfill & \text{and}\hfill & \underset{x\to a}{\lim }h(x)=-\infty \hfill \end{array}$

This theorem holds if $x\to a$ is replaced by $x\to a^{+}$ or $x\to a^{-}\text{.}$

Tutorial 2: The steps we must follow to conclude that a limit does not exist.

Example 3.47. The limit

${\displaystyle \underset{x\to 0^{-}}{\lim }\frac{x-5}{\sqrt{x+9}-3}}$

is of the form ${\displaystyle \underset{x\to 0^{-}}{\lim }\left(x-5\right)\frac{1}{\sqrt{x+9}-3}\text{.}}$

1. ${\displaystyle \underset{x\to 0^{-}}{\lim }\frac{1}{\sqrt{x+9}-3}=-\infty }$ by Example 3.43
2. ${\displaystyle \underset{x\to 0^{-}}{\lim }x-5=-5<0}$ by Theorem 3.14

Hence, by Theorem 3.18(b), ${\displaystyle \underset{x\to 0^{-}}{\lim }\frac{x-5}{\sqrt{x+9}-3}=\infty }\text{.}$

Note: An algebraic manipulation of the function $\color{#384877}{f(x) = \dfrac{x - 5}{\sqrt{x + 9} -3}}$ is not a better approach than the one we use here, try it.

Example 3.48. For the limit

${\displaystyle \underset{x\to 0^{-}}{\lim }\frac{x^2+1}{x^{2}sinx}=\underset{x\to 0^{-}}{\lim }(x^2+1)\frac{1}{x^2\sin x}\text{,}}$

we have $x^2\sin x<0$, for $x\to 0^{-}\text{.}$

Furthermore, by Theorem 3.14,

$\underset{x\to 0^{-}}{\lim }{x}^2+1=1>0\text{and}\underset{x\to 0^{-}}{\lim }{x}^2\sin x=0\text{.}$

Therefore, by Theorem 3.17,

${\displaystyle \underset{x\to 0^{-}}{\lim }\frac{1}{x^2\sin x}=-\infty \text{.}}$

We conclude, by Theorem 3.18, that

${\displaystyle \underset{x\to 0^{-}}{\lim }\frac{x^2+1}{x^2\sin x}=-\infty \text{.}}$

Example 3.49. Let us evaluate the limit ${\displaystyle \underset{x\to 2}{\lim }\frac{\sqrt{x+3}}{{\left(x-2\right)}^2}}\text{.}$

1. $\underset{x\to 2}{\lim }\sqrt{x+3}=\sqrt{5}$ by Theorem 3.14.

2. ${\displaystyle \underset{x\to 2}{\lim }\frac{1}{{\left(x-2\right)}^2}=\infty }$ since ${\displaystyle \underset{x\to 2}{\lim }{\left(x-2\right)}^2=0}$ and ${\left(x-2\right)}^2>0$ for $x\to 2$ (Theorem 3.17).

Hence, by Theorem 3.18,

${\displaystyle \underset{x\to 2}{\lim }\frac{\sqrt{x+3}}{{\left(x-2\right)}^2}=\underset{x\to 2}{\lim }\sqrt{x+3}\frac{1}{{\left(x-2\right)}^2}=\infty \text{.}}$

Example 3.50. $\underset{x\to \pi ∕2^{+}}{\lim }\left(x^2-3\right)\sec x=\infty$

because

1. $\underset{x\to \pi /2^{+}}{\lim }{x}^2-3=\frac{{\pi }^2}{4}-3<0$ by Theorem 3.14.

2. ${\displaystyle \underset{x\to \pi /2^{+}}{\lim }\sec x=-\infty }$ by Theorem 3.17.

From the graphs of Examples 3.39 to 3.42, we can see that a vertical line $x=a$ is a vertical asymptote if a limit at $a$ is infinity.

To find vertical asymptotes, we take any of the first four limits listed in Definition 3.19 (since these limits always exist), at a point or points $a$ where the function is not defined. But note that, even when a function is not defined at a point $a$, the line $x=a$ may not be a vertical asymptote.

PowerPoint 3: For x close to α from the left, the values of the function f(x) grow without bound. Hence x = α is a vertical asymptote.

PowerPoint 4: For x close to α from the right, the values of the function f(x) grow without bound. Hence x = α is a vertical asymptote.

PowerPoint 5: For x close to α from the left, the values of the function f(x) decrease without bound. Hence x = α is a vertical asymptote.

PowerPoint 6: For x close to α from the right, the values of the function f(x) decrease without bound. Hence x = α is a vertical asymptote.

Example 3.51. Although the function

${\displaystyle f\left(h\right)=\frac{{\left(4+h\right)}^2-16}{h^2+h}}$

is not defined at $0$, the vertical line $x=0$ is not a vertical asymptote, because, as we showed in Example 3.35, the limit

${\displaystyle \underset{h\to 0}{\lim }\frac{{\left(4+h\right)}^2-16}{h^2+h}=8}$

is finite.

However, the function is not defined at $-1$ either. [Why?]

We have

${\displaystyle \underset{h\to -1^{+}}{\lim }\frac{{(4+h)}^2-16}{h^2+h}=\underset{h\to -1^{+}}{\lim }\frac{8+h}{h+1}=\infty }$

by Theorem 3.18, since

${\displaystyle \underset{h\to -1^{+}}{\lim }\frac{1}{h+1}=\infty \text{.}}$  [By which theorem?]

Finally,

$\underset{h\to -1^{+}}{\lim }8+h=7>0.$

Hence, for this function $x=-1$ is a vertical asymptote.

For the composition of functions, we have the following theorem.

Theorem 3.20. If

$\underset{x\to a}{\lim }g\left(x\right)=b\text{and}\underset{x\to b}{\lim }f\left(x\right)=\pm \infty \text{,}$

then

$\underset{x\to a}{\lim }f\left(g\left(x\right)\right)=\pm \infty \text{.}$

The theorem holds if $x\to a$ is replaced by either $x\to a^{+}$ or $x\to a^{-}$; and $x\to b$ is replaced by either $x\to b^{+}$ or $x\to b^{-}\text{.}$

Theorem 3.20 says that if $u=g\left(x\right)$, then $u=g\left(x\right)\to b$ as $x\to a$, and

$\underset{x\to a}{\lim }f\left(g\left(x\right)\right)=\underset{u\to b}{\lim }f\left(u\right)=\pm \infty$

Example 3.52. The function in the limit

$\underset{x\to {\pi }^{+}}{\lim }\tan \left(x-{\displaystyle \frac{\pi }{2}}\right)$

is a composition of functions.

For $x\to {\pi }^{+}$, we have

${\displaystyle x-\frac{\pi }{2}>\frac{{\pi }^{+}}{2}\text{;}}$

hence,

${\displaystyle u=x-\frac{\pi }{2}\to \frac{{\pi }^{+}}{2}\text{as}x\to {\pi }^{+}\text{.}}$

Furthermore, by Theorem 3.20,

${\displaystyle \underset{x\to {\pi }^{+}}{\lim }\tan \left(x-\frac{\pi }{2}\right)=\underset{x\to {\pi ∕2}^{+}}{\lim }\tan u=-\infty }$

Observe that if $\underset{x\to a}{\lim }f\left(x\right)=\infty$ and $\underset{x\to a}{\lim }g\left(x\right)=\infty$, then for $x\to a$, both $f\left(x\right)$ and $g\left(x\right)$ increase without bound, and their sum and product also increase without bound.

Thus, $\underset{x\to a}{\lim }f\left(x\right)+g\left(x\right)=\infty$ and $\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=\infty \text{.}$

Moreover, for any nonzero number $c$, we know, by Theorem 3.18, that

$\underset{x\to a}{\lim }cf\left(x\right)=\pm \infty \text{if}c>0\text{or}c<0\text{.}$

This fact explains, in part, the following theorem.

Theorem 3.21.

1. If $\underset{x\to a}{\lim }f\left(x\right)=\infty$ and $\underset{x\to a}{\lim }g\left(x\right)=\infty$ and $c$ is a nonzero constant, then

   $\begin{array}{ll}\underset{x\to a}{\lim }f\left(x\right)+g\left(x\right)=\infty \text{,}\hfill & \underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=\infty \text{ and}\hfill \\ \underset{x\to a}{\lim }cf\left(x\right)=\pm \infty \text{, if }\hfill & c>0\text{ or }c<0.\hfill \end{array}$

2. If $\underset{x\to a}{\lim }f\left(x\right)=-\infty$ and $\underset{x\to a}{\lim }g\left(x\right)=-\infty$ and $c$ is a nonzero constant, then

   $\begin{array}{ll}\underset{x\to a}{\lim }f\left(x\right)+g\left(x\right)=-\infty \text{,}\hfill & \underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=\infty \text{ and}\hfill \\ \underset{x\to a}{\lim }cf\left(x\right)=\pm \infty \text{, if }\hfill & c>0\text{ or }c<0.\hfill \end{array}$

3. If $\underset{x\to a}{\lim }f\left(x\right)=\infty$ and $\underset{x\to a}{\lim }g\left(x\right)=-\infty$, then $\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=-\infty$

Example 3.53. Let us evaluate the limit

$\underset{x\to 0^{+}}{\lim }\csc x+\cot x\text{.}$

We know that

$\underset{x\to 0^{+}}{\lim }\csc x=\infty \text{and}\underset{x\to 0^{+}}{\lim }cotx=\infty \text{;}$

so, by Theorem 3.21, we conclude that

$\underset{x\to 0^{+}}{\lim }\csc x+\cot x=\infty \text{.}$

Warning: Theorem 3.21 does not say that

$\underset{x\to a}{\lim }f\left(x\right)+g\left(x\right)=\infty +\infty =\infty \text{.}$

This expression is incorrect because $\infty$ is a symbol, and it does not make sense to add symbols. Instead, Theorem 3.21 states that if two functions increase without bound, then their sum also increases without bound.

Example 3.54. By Theorem 3.21,

${\displaystyle \underset{x\to {\pi ∕2}^{-}}{\lim }-\frac{\sec x}{3}=-\infty \text{,}}$

because $-1∕3<0$ and $\underset{x\to {\pi ∕2}^{-}}{\lim }\sec x=\infty \text{.}$

Warning: Theorem 3.21 does not say that

$\underset{x\to a}{\lim }cf\left(x\right)=c\infty \text{.}$

This expression is incorrect because $\infty$ is a symbol, and it does not make sense to multiply a number by a symbol. Instead, Theorem 3.21 states that if a function increases without bound then the product of the multiplication of the function by a positive constant also increases without bound.

When working with infinite limits, we must be careful to apply theorems correctly, and not to infer more than they indicate. For instance, if $\underset{x\to a}{\lim }f\left(x\right)=\infty$ and $\underset{x\to a}{\lim }g\left(x\right)=-\infty$, it is not true that $\underset{x\to a}{\lim }f\left(x\right)+g\left(x\right)=0\text{.}$