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2.4 Infinite Limits

\[ f(x) = \frac{1}{(x-2)^2} \]\[ \lim_{x \to 2} \frac{1}{(x-2)^2} = ? \]

\( x \)	1.99	1.999	1.9999	2	2.0001	2.001	2.01
\( \frac{1}{(x-2)^2} \)	10,000	1,000,000	100,000,000	X	100,000,000	1,000,000	10,000

It appears as \( x \to 2 \) from either side, \( \frac{1}{(x-2)^2} \) keeps getting bigger, the closer we get to \( x = 2 \), the bigger \( \frac{1}{(x-2)^2} \) is.

Since \( \frac{1}{(x-2)^2} \) as \( x \to 2 \) grows without bound,

\[ \text{we say } \lim_{x \to 2} \frac{1}{(x-2)^2} = \infty \quad \text{(infinite limit } \to \text{ when limit is } \pm \infty) \]

we also see: \[ \lim_{x \to 2^+} \frac{1}{(x-2)^2} = \infty \quad , \quad \lim_{x \to 2^-} \frac{1}{(x-2)^2} = \infty \]

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graph:

vertical asymptote at \( x = 2 \)

graph goes to \( \infty \) or \( -\infty \) on either side

vertical asymptote: numerator \( \neq 0 \) while denominator \( = 0 \)

example:

\[ \lim_{x \to 2} \frac{1}{x-2} \]

\( x \)	1.99	1.999	1.9999	2	2.0001	2.001	2.01
\( \frac{1}{x-2} \)	-100	-1000	-10,000	X	10,000	1000	100

we see:

\( \lim_{x \to 2^-} \frac{1}{x-2} = -\infty \) (keeps getting bigger but negative)
\( \lim_{x \to 2^+} \frac{1}{x-2} = \infty \) (keeps increasing remains positive)
\( \lim_{x \to 2} \frac{1}{x-2} \text{ DNE} \) because the one-sided limits do not match

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Graph Analysis

The following graph illustrates the behavior of a function near a vertical asymptote.

Figure: Graph of a function with a vertical asymptote at x=2, showing curves approaching infinity and negative infinity.

In practice, we can just check using numbers close to the target on either side.

\[ \frac{\text{non zero}}{\text{small number}} \rightarrow \infty \text{ or } -\infty \]

depending on the sign of the ratio

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Example

\[ \lim_{x \to 5^+} \frac{x^2 - 7x + 12}{(x - 5)^2} \]

\( x \to 5^+ \rightarrow \) pick a number slightly bigger than 5.

For example, \( x = 5.0001 \).

Plug into \( \frac{x^2 - 7x + 12}{(x - 5)^2} \):

\[ = \frac{(5.0001)^2 - 7(5.0001) + 12}{(5.0001 - 5)^2} \approx \frac{\text{roughly 2}}{\text{very small positive number}} \rightarrow \text{big positive #} \]

Ratio is positive

So, \[ \lim_{x \to 5^+} \frac{x^2 - 7x + 12}{(x - 5)^2} = \infty \]

Follow-up:

\[ \lim_{x \to 5^-} \frac{x^2 - 7x + 12}{(x - 5)^2} \]

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Limits and One-Sided Limits

Pick slightly smaller than 5: \( x = 4.999 \)

Plug into:

\[ \frac{x^2 - 7x + 12}{(x-5)^2} \approx \frac{\text{roughly } 2}{\text{positive small #}} \]

ratio is positive

So, \[ \lim_{x \to 5^-} \frac{x^2 - 7x + 12}{(x-5)^2} = \infty \]

Since the one-sided limits match, \[ \lim_{x \to 5} \frac{x^2 - 7x + 12}{(x-5)^2} = \infty \]

Example

\[ \lim_{x \to 4^+} \frac{x-7}{\sqrt{x-4}} \]

\( x \to 4^+ \) : \( x \) is something something bigger than 4

Pick \( x = 4.01 \)

then

\[ \frac{x-7}{\sqrt{x-4}} = \frac{\text{roughly } -3}{\sqrt{0.01}} \to \frac{\text{roughly } -3}{\text{small positive #}} \]

ratio is negative

So, \[ \lim_{x \to 4^+} \frac{x-7}{\sqrt{x-4}} = -\infty \]

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What about \[ \lim_{x \to 4^-} \frac{x-7}{\sqrt{x-4}} \text{ ?} \]

\( x \to 4^- \) means pick something smaller than 4

Pick \( x = 3.99 \)

\[ \frac{x-7}{\sqrt{x-4}} = \frac{\text{roughly } -3}{\sqrt{-0.01}} \]

\( \sqrt{-0.01} \) is undefined

no ratio to talk about

the entire ratio is undefined

All we can say is \[ \lim_{x \to 4^-} \frac{x-7}{\sqrt{x-4}} \text{ DNE (does not exist)} \]

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Example

\[ \lim_{x \to 4} \frac{x^2 - 9x + 14}{x^2 - 6x + 8} \]

looks like we can factor things and simplify

do that first

\[ \frac{x^2 - 9x + 14}{x^2 - 6x + 8} = \frac{(x - 7)(x - 2)}{(x - 4)(x - 2)} \]

\[ = \lim_{x \to 4} \frac{x - 7}{x - 4} \]

Evaluating the Limit

find \( \lim_{x \to 4^+} \frac{x - 7}{x - 4} \)

choose \( x = 4.001 \)

\[ \frac{x - 7}{x - 4} \to \frac{\text{roughly } -3}{\text{small positive #}} \to -\infty \]

find \( \lim_{x \to 4^-} \frac{x - 7}{x - 4} = \infty \)

\[ \lim_{x \to 4} \frac{x - 7}{x - 4} \text{ DNE} \]