Wize High School Grade 12 Calculus Textbook > Rate of Change

Definition of Limits
Left-hand limit: If 𝑓(𝑥) gets close to the finite number 𝐿 as 𝑥 approaches 𝑎 from the left, then lim⁡x→a−f(x)=L\displaystyle\lim_{x\to a^-}f\left(x\right)=Lx→a−lim​f(x)=L.

Right-hand limit: If 𝑓(𝑥) gets close to the finite number 𝐿 as 𝑥 approaches 𝑎 from the right, then lim⁡x→a+f(x)=L\displaystyle\lim_{x\to a^+}f\left(x\right)=Lx→a+lim​f(x)=L.

Limit of a function: If these two limits are equal, then the limit of 𝑓(𝑥) as 𝑥 approaches 𝑎 is defined and equals 𝐿
lim⁡x→a−f(x)=lim⁡x→a+f(x) ↔ lim⁡x→af(x)=L\displaystyle\lim_{x\to a^-}f\left(x\right)=\lim_{x\to a^+}f\left(x\right)\ \leftrightarrow\ \lim_{x\to a}f\left(x\right)=Lx→a−lim​f(x)=x→a+lim​f(x) ↔ x→alim​f(x)=L 
Note: If the left and right hand limits do not equal each other, the limit does not exist (DNE).

Watch Out!
𝑓(𝑥) does not have to equal 𝐿, it doesn't even have to be defined at 𝑥 = 𝑎, just at values really close to 𝑎. 

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Example
The following is the graph of the function f(x)f\left(x\right)f(x)
Determine the following limits:

a. lim⁡x→0f(x)=\displaystyle \lim_{x\to0}f\left(x\right)=x→0lim​f(x)= 1

Because the left and right hand limits both approach 1.

b. i)lim⁡x→2−f(x)=\displaystyle \lim_{x\to2^-}f\left(x\right)=x→2−lim​f(x)=4
            ii) lim⁡x→2+f(x)=\displaystyle \lim_{x\to2^+}f\left(x\right)=x→2+lim​f(x)=4
            iii) lim⁡x→2f(x)=\displaystyle \lim_{x\to2}f\left(x\right)=x→2lim​f(x)=4

The left and right hand limits both approach 4, so even though the function value f(4)=−2f\left(4\right)=-2f(4)=−2, the limit actually equals 4

c. i) lim⁡x→5−f(x)=\displaystyle \lim_{x\to5^-}f\left(x\right)=x→5−lim​f(x)= -1
         ii) lim⁡x→5+f(x)=\displaystyle \lim_{x\to5^+}f\left(x\right)=x→5+lim​f(x)= -3
         iii) lim⁡x→5f(x)=\displaystyle \lim_{x\to5}f\left(x\right)=x→5lim​f(x)=DNE
 

Since the left hand limit approaches -1 and the right hand limit approaches -3, the limit does not exist because the values don't match up.

d. i) lim⁡x→8−f(x)=\displaystyle \lim_{x\to8^-}f\left(x\right)=x→8−lim​f(x)=-∞ (DNE)
          ii) lim⁡x→8+f(x)=\displaystyle \lim_{x\to8^+}f\left(x\right)=x→8+lim​f(x)=-∞ (DNE)
         iii) lim⁡x→8f(x)=\displaystyle \lim_{x\to8}f\left(x\right)=x→8lim​f(x)=-∞ (DNE)

Since the left and right hand limits don't approach a single numerical value, it does not exist.

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Example: Limits From Graphs
The population (in thousands) of a certain species of wild cats at time ttt (in years) is modelled by the function
p(t)={10,if 0≤t<52t2−40,if 5≤t<840t+60,if 8≤tp\left(t\right)=\begin{cases}
10,&\text{if }0\le t<5\\
2t^2-40,&\text{if }5\le t<8\\
\frac{40}{t}+60,&\text{if }8\le t
\end{cases}p(t)=⎩⎨⎧​10,2t2−40,t40​+60,​if 0≤t<5if 5≤t<8if 8≤t​

a) Sketch the graph of p(t)p(t)p(t).
b) Evaluate p(0)p(0)p(0), p(3)p(3)p(3), p(5)p(5)p(5), p(8)p(8)p(8), and p(10)p(10)p(10).
c) Evaluate lim⁡t→5−p(t)\displaystyle \lim_{t\to5^-}p\left(t\right)t→5−lim​p(t), lim⁡t→5+p(t)\displaystyle \lim_{t\to5^+}p\left(t\right)t→5+lim​p(t), and lim⁡t→5p(t)\displaystyle \lim_{t\to5}p\left(t\right)t→5lim​p(t).
d) Evaluate lim⁡t→6−p(t)\displaystyle \lim_{t\to6^-}p\left(t\right)t→6−lim​p(t), lim⁡t→6+p(t)\displaystyle \lim_{t\to6^+}p\left(t\right)t→6+lim​p(t), and lim⁡t→6p(t)\displaystyle \lim_{t\to6}p\left(t\right)t→6lim​p(t).
e) Evaluate lim⁡t→8−p(t)\displaystyle \lim_{t\to8^-}p\left(t\right)t→8−lim​p(t), lim⁡t→8+p(t)\displaystyle \lim_{t\to8^+}p\left(t\right)t→8+lim​p(t), and lim⁡t→8p(t)\displaystyle \lim_{t\to8}p\left(t\right)t→8lim​p(t).
f) At time t=8t=8t=8, there was a predator that reduced the population significantly. Determine the number of wild cats that were killed by this predator at that time.

Part a) 
Part b)
p(0)=10p\left(0\right)=10p(0)=10
p(3)=10p\left(3\right)=10p(3)=10
p(5)=2(5)2−40=10p\left(5\right)=2\left(5\right)^2-40=10p(5)=2(5)2−40=10
p(8)=408+60=65p\left(8\right)=\frac{40}{8}+60=65p(8)=840​+60=65
p(10)=4010+60=64p\left(10\right)=\frac{40}{10}+60=64p(10)=1040​+60=64

Part c)
lim⁡t→5−p(t)=lim⁡t→5−10=10\displaystyle \lim_{t\to5^-}p\left(t\right)=\displaystyle \lim_{t\to5^-}10=10t→5−lim​p(t)=t→5−lim​10=10
lim⁡t→5+p(t)=lim⁡t→5+2t2−40=10\displaystyle \lim_{t\to5^+}p\left(t\right)=\displaystyle \lim_{t\to5^+}2t^2-40=10t→5+lim​p(t)=t→5+lim​2t2−40=10
lim⁡t→5p(t)=10\displaystyle \lim_{t\to5}p\left(t\right)=10t→5lim​p(t)=10

Part d) 
lim⁡t→6−p(t)=lim⁡t→6−2t2−40=32\displaystyle \lim_{t\to6^-}p\left(t\right)=\displaystyle \lim_{t\to6^-}2t^2-40=32t→6−lim​p(t)=t→6−lim​2t2−40=32
lim⁡t→6+p(t)=lim⁡t→6+2t2−40=32\displaystyle \lim_{t\to6^+}p\left(t\right)=\displaystyle \lim_{t\to6^+}2t^2-40=32t→6+lim​p(t)=t→6+lim​2t2−40=32
lim⁡t→6p(t)=32\displaystyle \lim_{t\to6}p\left(t\right)=32t→6lim​p(t)=32

Part e)
lim⁡t→8−p(t)=lim⁡t→8−2t2−40=88\displaystyle \lim_{t\to8^-}p\left(t\right)=\displaystyle \lim_{t\to8^-}2t^2-40=88t→8−lim​p(t)=t→8−lim​2t2−40=88
lim⁡t→8+p(t)=lim⁡t→8+40t+60=65\displaystyle \lim_{t\to8^+}p\left(t\right)=\displaystyle \lim_{t\to8^+}\frac{40}{t}+60=65t→8+lim​p(t)=t→8+lim​t40​+60=65
lim⁡t→8p(t)=DNE\displaystyle \lim_{t\to8}p\left(t\right)=DNEt→8lim​p(t)=DNE

Part f) 
Since lim⁡t→8−p(t)=88\displaystyle \lim_{t\to8^-}p\left(t\right)=88t→8−lim​p(t)=88, we know that right before the predator killed a portion of the wild cats, there were 88 thousand wild cats.

Sincelim⁡t→8+p(t)=65\displaystyle \lim_{t\to8^+}p\left(t\right)=65t→8+lim​p(t)=65, we know that after the predator attack, there were only 65 thousand wild cats.

Therefore, the predator killed 23 thousand wild cats at the beginning of year 8.


*BONUS*
Q: What is the population of this species of wild cats as time goes on? (i.e. 100 years from now, 10000 years from now, etc.)
A: As t→∞t\to\inftyt→∞, the population will follow the function 40t+60\frac{40}{t}+60t40​+60. So, as ttt gets really large, 40t\frac{40}{t}t40​ gets really small, until it's practically 0. So the population in the long run as time goes on will approach 60,000 wild cats.

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Limits Properties
Suppose that lim⁡x→af(x)\displaystyle\lim_{x\to a}f\left(x\right)x→alim​f(x) and lim⁡x→ag(x)\displaystyle\lim_{x\to a}g\left(x\right)x→alim​g(x) both exist.
Constant Multiplelim⁡x→a cf(x)=c lim⁡x→af(x)Limit Sum/Differencelim⁡x→a[f(x)±g(x)]=lim⁡x→af(x)±lim⁡x→ag(x)Limit Productlim⁡x→a[f(x)g(x)]=lim⁡x→af(x)×lim⁡x→ag(x)Limit Quotientlim⁡x→af(x)g(x)=lim⁡x→af(x)lim⁡x→ag(x)Limit Powerlim⁡x→a[f(x)]n=[lim⁡x→af(x)]n\begin{array}{|l|l|}
\hline\\
\text{Constant Multiple}&\displaystyle\lim_{x\to a}\ cf\left(x\right)=c\ \lim_{x\to a}f\left(x\right)\\\\

\hline\\
\text{Limit Sum/Difference}&\displaystyle\lim_{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim_{x\to a}f\left(x\right)\pm\lim_{x\to a}g\left(x\right)\\\\

\hline\\
\text{Limit Product}&\displaystyle\lim_{x\to a}[f\left(x\right)g\left(x\right)]=\lim_{x\to a}f\left(x\right)\times\lim_{x\to a}g\left(x\right)\\\\

\hline\\
\text{Limit Quotient}&\displaystyle\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=
\frac{\displaystyle\lim_{x\to a}f\left(x\right)}{\displaystyle\lim_{x\to a}g\left(x\right)}\\\\

\hline\\
\text{Limit Power}&\displaystyle\lim_{x\to a}\left[f\left(x\right)\right]^n=\left[\lim_{x\to a}f\left(x\right)\right]^n\\\\
\hline
\end{array}Constant MultipleLimit Sum/DifferenceLimit ProductLimit QuotientLimit Power​x→alim​ cf(x)=c x→alim​f(x)x→alim​[f(x)±g(x)]=x→alim​f(x)±x→alim​g(x)x→alim​[f(x)g(x)]=x→alim​f(x)×x→alim​g(x)x→alim​g(x)f(x)​=x→alim​g(x)x→alim​f(x)​x→alim​[f(x)]n=[x→alim​f(x)]n​​

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Direct Substitution
For any real number aaa:
lim⁡x→ak=k\displaystyle \lim_{x\to a}k=kx→alim​k=k for any constant kkk
lim⁡x→ax=a\displaystyle \lim_{x\to a}x=ax→alim​x=a
Meaning: Substitute the value that xxx approaches directly into the expression → if you get a number, that's the limit!

Wize Tip
ALWAYS TRY DIRECTION SUBSTITUTION FIRST!

Watch Out!
If you end up with 0\sqrt 00​, check the left and right-hand limit to make sure the limit exists

Watch Out!
If you end up with 00\frac{0}{0}00​, this is called an indeterminate form → have to manipulate the expression algebraically to evaluate the limit.

Other indeterminate forms (some of these will be covered in later chapters, some will not be covered in this course):
∞∞, 0×∞, ∞−∞, 1∞, ∞0, 00\frac{\infty}{\infty},\ 0\times\infty,\ \infty-\infty,\ 1^{\infty},\ \infty^0,\ 0^0∞∞​, 0×∞, ∞−∞, 1∞, ∞0, 00

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Example: Evaluating Limits
Evaluate the following limits:
a) lim⁡x→3 2x2+1x\displaystyle \lim_{x\to3}\ \frac{2x^2+1}{x}x→3lim​ x2x2+1​

lim⁡x→3 2x2+1x=2(3)2+13=193\displaystyle \lim_{x\to3}\ \frac{2x^2+1}{x}=\frac{2\left(3\right)^2+1}{3}=\frac{19}{3}x→3lim​ x2x2+1​=32(3)2+1​=319​

b) lim⁡x→−23x+8\displaystyle \lim_{x\to-2}\sqrt{3x+8}x→−2lim​3x+8​

lim⁡x→−23x+8=3(−2)+8=2\displaystyle \lim_{x\to-2}\sqrt{3x+8}=\sqrt{3(-2)+8}=\sqrt {2}x→−2lim​3x+8​=3(−2)+8​=2​

c) lim⁡x→−3∣x2+x−10∣\displaystyle \lim_{x\to-3}\left|x^2+x-10\right|x→−3lim​​x2+x−10​

lim⁡x→−3∣x2+x−10∣=∣(−3)2+(−3)−10∣=∣−4∣=4\displaystyle \lim_{x\to-3}\left|x^2+x-10\right|=|(-3)^2+(-3)-10|=|-4|=4x→−3lim​​x2+x−10​=∣(−3)2+(−3)−10∣=∣−4∣=4

d) lim⁡x→1x−1\displaystyle \lim_{x\to1}\sqrt{x-1}x→1lim​x−1​ (*tricky*)

If we do a direct substitution, we get lim⁡x→1x−1=1−1=0\displaystyle \lim_{x\to1}\sqrt{x-1}=\sqrt {1-1}=\sqrt 0x→1lim​x−1​=1−1​=0​
We know that 0\sqrt 00​ is defined, however, − number\sqrt {-\text{ number}}− number​ is not defined.
Let's examine both left and right hand limits:
lim⁡x→1+x−1=1+−1=0+=0\displaystyle \lim_{x\to1^+}\sqrt{x-1}=\sqrt{1^+-1}=\sqrt{0^+}=0x→1+lim​x−1​=1+−1​=0+​=0 (this is the square root of a very small positive value)
lim⁡x→1−x−1=1−−1=0−=DNE\displaystyle \lim_{x\to1^-}\sqrt{x-1}=\sqrt{1^--1}=\sqrt{0^-}=DNEx→1−lim​x−1​=1−−1​=0−​=DNE (this is the square root of a very small negative value)
Therefore, the limit does not exist.

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e) f(x)={∣x∣,x<02x2−1,x≥0f\left(x\right)=
\begin{cases}
|x|,&x<0\\
2x^2-1,&x\ge0
\end{cases}f(x)={∣x∣,2x2−1,​x<0x≥0​
i) lim⁡x→−1f(x)\displaystyle \lim_{x\to-1}f\left(x\right)x→−1lim​f(x)

lim⁡x→−1f(x)=lim⁡x→−1∣x∣=∣−1∣=1\displaystyle \lim_{x\to-1}f\left(x\right)=\lim_{x\to-1}|x|=|-1|=1x→−1lim​f(x)=x→−1lim​∣x∣=∣−1∣=1

ii) lim⁡x→1f(x)\displaystyle \lim_{x\to1}f\left(x\right)x→1lim​f(x)

lim⁡x→1f(x)=lim⁡x→−12x2−1=2(1)2−1=1\displaystyle \lim_{x\to1}f\left(x\right)=\lim_{x\to-1}2x^2-1=2(1)^2-1=1x→1lim​f(x)=x→−1lim​2x2−1=2(1)2−1=1

iii) lim⁡x→0f(x)\displaystyle \lim_{x\to0}f\left(x\right)x→0lim​f(x)

Since the function switches at the point x=0x=0x=0, we need to examine the left and right hand limits:
lim⁡x→0−f(x)=lim⁡x→0−∣x∣=∣0−∣=0\displaystyle \lim_{x\to0^-}f\left(x\right)=\lim_{x\to0^-}\left|x\right|=\left|0^-\right|=0x→0−lim​f(x)=x→0−lim​∣x∣=​0−​=0
lim⁡x→0+f(x)=lim⁡x→0+2x2−1=2(0+)2−1=−1\displaystyle \lim_{x\to0^+}f\left(x\right)=\lim_{x\to0^+}2x^2-1=2\left(0^+\right)^2-1=-1x→0+lim​f(x)=x→0+lim​2x2−1=2(0+)2−1=−1

Practice: Evaluating Limits
Evaluate the limit lim⁡x→3 ∣x2−10∣−x+2x\displaystyle \lim_{x\to3}\ \frac{\left|x^2-10\right|-x+2}{x}x→3lim​ x​x2−10​−x+2​

Enter your answer in exact value or enter DNE if the limit does not exist

I don't know

Practice: Evaluating Limits
Given the following function, evaluate the limits below.
f(x)={−3x+5if  x<−1−10if  x=−19−x2if  −1<x≤21−∣2x−3∣if  x>2f\left(x\right)=
\begin{cases}
-3x+5 & \text{if } \  x<-1 \\
-10 & \text{if }\ x=-1\\
9-x^2 & \text{if }\  -1< x\le2\\
1-|2x-3| & \text{if } \ x>2
\end{cases}f(x)=⎩⎨⎧​−3x+5−109−x21−∣2x−3∣​if  x<−1if  x=−1if  −1<x≤2if  x>2​

Enter DNE if the limit does not exist.

a. i)

\displaystyle\lim_{x\to-1^-}f\left(x\right)=

a. ii)

\displaystyle\lim_{x\to-1^+}f\left(x\right)=

a. iii)

\displaystyle\lim_{x\to-1}f\left(x\right)=

b. i)

\displaystyle\lim_{x\to2^-}f\left(x\right)=

b. ii)

\displaystyle\lim_{x\to2^+}f\left(x\right)=

b. iii)

\displaystyle\lim_{x\to2}f\left(x\right)=

c. i)

\displaystyle\lim_{x\to3^-}f\left(x\right)=

c. ii)

\displaystyle\lim_{x\to3^+}f\left(x\right)=

c. iii)

\displaystyle\lim_{x\to3}f\left(x\right)=

I don't know

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Example: Evaluating Limits
If lim⁡x→−2 f(x)=3\displaystyle\lim_{x\to-2}\ f\left(x\right)=3x→−2lim​ f(x)=3 and lim⁡x→−2 g(x)f(x)=5\displaystyle\lim_{x\to-2}\ \frac{g\left(x\right)}{f\left(x\right)}=5x→−2lim​ f(x)g(x)​=5, then lim⁡x→−2 g(x)x+4= \displaystyle\lim_{x\to-2}\ \frac{g\left(x\right)}{x+4}=\ x→−2lim​ x+4g(x)​= ?

Using what we know, we need to create the expression g(x)x+4\frac{g\left(x\right)}{x+4}x+4g(x)​:
f(x)⋅g(x)f(x)⋅1x+4=g(x)x+4f\left(x\right)\cdot\frac{g\left(x\right)}{f\left(x\right)}\cdot\frac{1}{x+4}=\frac{g\left(x\right)}{x+4}f(x)⋅f(x)g(x)​⋅x+41​=x+4g(x)​

So, we know that
lim⁡x→−2 g(x)x+4\displaystyle\lim_{x\to-2}\ \frac{g\left(x\right)}{x+4}x→−2lim​ x+4g(x)​
= lim⁡x→−2 [f(x)⋅g(x)f(x)⋅1x+4]\displaystyle=\ 

\lim_{x\to-2}\ \left[f(x)\cdot\frac{g(x)}{f(x)}\cdot\frac{1}{x+4}\right]= x→−2lim​ [f(x)⋅f(x)g(x)​⋅x+41​]
=lim⁡x→−2[f(x)]⋅lim⁡x→−2[g(x)f(x)]⋅lim⁡x→−2[1x+4]=\displaystyle\lim_{x\to-2}\left[f\left(x\right)\right]\cdot\lim_{x\to-2}\left[\frac{g\left(x\right)}{f\left(x\right)}\right]\cdot\lim_{x\to-2}\left[\frac{1}{x+4}\right]=x→−2lim​[f(x)]⋅x→−2lim​[f(x)g(x)​]⋅x→−2lim​[x+41​]
=3⋅5⋅1(−2)+4=3\cdot5\cdot\frac{1}{\left(-2\right)+4}=3⋅5⋅(−2)+41​
=152=\frac{15}{2}=215​

Practice: Evaluating Limits
If lim⁡x→1f(x)=−2\displaystyle \lim_{x\to1}f\left(x\right)=-2x→1lim​f(x)=−2, evaluate the limit  lim⁡x→1 2f(x)+5[f(x)]3+x\displaystyle \lim_{x\to1}\ \frac{\sqrt{2f\left(x\right)+5}}{\left[f\left(x\right)\right]^3+x}x→1lim​ [f(x)]3+x2f(x)+5​​.

Enter the exact value of the limit, or DNE if the limit does not exist

I don't know

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This concept will be revisited in the Curve Sketching chapter.
Infinite Limits
After you try direct substitution, if you get non-zero number0\displaystyle \orange{\frac{\text{non-zero number}}{0}}0non-zero number​, then 
  a. if the overall value is positive, the limit equals +∞+\infty+∞ 
  b. if the overall value is negative, the limit equals −∞-\infty−∞
  c. if you can't know for sure if the overall value is positive or negative, the limit DNE

Example
1. lim⁡x→2+ 12−x=\displaystyle\lim_{x\to2^+}\ \frac{1}{2-x}=x→2+lim​ 2−x1​=-∞ (DNE)

Direct sub gives us 10\frac{1}{0}01​. If we substitute 2+2^+2+ into the expression, the denominator will be slightly negative. Since the numerator is positive 1, the overall value will be negative. Therefore, the limit is −∞-\infty−∞ (DNE)

2. lim⁡x→2 −e∣x−2∣=\displaystyle\lim_{x\to2}\ -\frac{e}{|x-2|}=x→2lim​ −∣x−2∣e​=-∞ (DNE)

Direct sub gives us −e0-\frac{e}{0}−0e​. It doesn't matter if we substitute 2−2^-2− or 2+2^+2+ into the expression, the deonominator will be positive, and the overall value will be negative. Therefore, the limit is −∞-\infty−∞ (DNE)

3. lim⁡x→π 1π−x=\displaystyle\lim_{x\to\pi}\ \frac{1}{\pi-x}=x→πlim​ π−x1​=DNE

Direct sub gives us cos⁡π0=−10\frac{\cos\pi}{0}=-\frac{1}{0}0cosπ​=−01​. If we substitute π−\pi^-π− into the expression, the denominator will be slightly positive, and the overall value will be negative. If we substitute π+\pi^+π+ into the expression, the denominator will be slightly negative, and the overall value will be positive. Therefore, the limit DNE.

Did You Know?
If lim⁡x→af(x)=±∞\displaystyle\lim_{x\to a}f\left(x\right)=\pm\inftyx→alim​f(x)=±∞, then f(x)f\left(x\right)f(x) has a vertical asymptote at x=ax=ax=a.

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This concept will be revisited in the Curve Sketching chapter.
Limits at Infinity & Horizontal Asymptotes
If we have lim⁡x→±∞f(x)\orange{\displaystyle\lim_{x\to\pm\infty}f\left(x\right)}x→±∞lim​f(x), substitute a really large positive or negative number in for xxx, then
  a. if we get non-zero number±∞\displaystyle \frac{\text{non-zero number}}{\pm\infty}±∞non-zero number​, then the limit is 000
  b. if we get ∞+∞\infty+\infty∞+∞, then the limit is ∞\infty∞
  c. if we get −∞−∞-\infty-\infty−∞−∞, then the limit is −∞-\infty−∞
  d. if we get ±∞non-zero number\displaystyle \frac{\pm\infty}{\text{non-zero number}}non-zero number±∞​, then the limit is ±∞\pm\infty±∞

Example
1. lim⁡x→−∞ 24−2x+x2=\displaystyle \lim_{x\to-\infty}\ \frac{2}{4-2x+x^2}=x→−∞lim​ 4−2x+x22​= 0 

Direct sub gives us 24−2(−∞)+(−∞)2=24+∞+∞=2∞=0\frac{2}{4-2\left(-\infty\right)+\left(-\infty\right)^2}=\frac{2}{4+\infty+\infty}=\frac{2}{\infty}=04−2(−∞)+(−∞)22​=4+∞+∞2​=∞2​=0

2. lim⁡x→∞ x+3+x−3=\displaystyle \lim_{x\to\infty}\ \sqrt{x+3}+\sqrt{x-3}=x→∞lim​ x+3​+x−3​=∞ (DNE)

Direct sub gives us ∞+3+∞−3=∞+∞=∞\sqrt{\infty+3}+\sqrt{\infty-3}=\infty+\infty=\infty∞+3​+∞−3​=∞+∞=∞ (DNE)

Did You Know?
If lim⁡x→∞f(x)=a\displaystyle\lim_{x\to \infty }f\left(x\right)=ax→∞lim​f(x)=a or lim⁡x→−∞f(x)=a\displaystyle\lim_{x\to -\infty }f\left(x\right)=ax→−∞lim​f(x)=a, then f(x)f\left(x\right)f(x) has a horizontal asymptote at y=ay=ay=a.

Watch Out!
If you get the result  ∞−∞\infty-\infty∞−∞ or ±∞±∞\frac{\pm\infty}{\pm\infty}±∞±∞​, we don't actually know what happens!

We will learn how to evaluate these types of limits in later chapters