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Expand Up @@ -34,7 +34,7 @@ $$lim_{{x \to 3}} \frac{{x^2 - 9}}{{x - 3}}$$

<br>

* **Simplified Form:** The numerator $\large x^2 - 9$, can be factored as ( (x + 3)(x - 3) ), which simplifies the expression to:
* **Simplified Form:** The numerator $$\large x^2 - 9$$, can be factored as (x + 3)(x - 3) ), which simplifies the expression to:

<br>

Expand All @@ -45,9 +45,9 @@ $$\\begin{align*}

<br>

* **Final Result:** Substituting ( x ) with 3, we get:
* [**Final Result:**]()

<br>
Substituting ( x ) with 3, we get:

$$\large 3 + 3 = 6$$

Expand Down Expand Up @@ -75,7 +75,9 @@ $$\large \lim_{{x \to -7}} (7 - x)
$$

<br>

* **Final Result:** When we substitute ( x ) with -7, the expression simplifies to:
7 - (-7) = 14
* [**Final Result:**]()

When we substitute ( x ) with -7, the expression simplifies to:
7 - (-7) = 14

<br>

Expand All @@ -85,65 +87,18 @@ $$\large \lim_{{x \to -7}} (7 - x)
$$

#

#### 1c) 8 88**Limit Expression:**


$$lim_{{\Large \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}}$$
















































### 1c) **Limit Expression:**


$$lim_{{\Large \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}}$$n

<br>

<!-- -- ---- -- --- -- --- Review Later -- ------ -->

### 1d) **Limit Expression:**


#### 1e) $$\lim_{{x \to 1}} \frac{{x^2 - 2x + 1}}{{x - 1}}$$
$$\lim_{{x \to 1}} \frac{{x^2 - 2x + 1}}{{x - 1}}$$

<br>

To calculate the limit, we can simplify the expression by factoring the numerator, which is a perfect square trinomial. Factoring (x^2 - 2x + 1), we get ((x - 1)(x - 1)). The denominator is already in factored form as (x - 1). Thus, the function simplifies to:

Expand All @@ -157,11 +112,17 @@ Since there are no more terms that depend on ( x ), this simplifies to:

$$\lim_{{x \to 1}} = x - 1 = 0$$

Result: The limit of the function as ( x ) approaches 1 is simply 0.
[Final Result:]()

The limit of the function as ( x ) approaches 1 is simply 0.

#

### 1f) $$\lim_{{x \to 2}} \frac{{x - 2}}{{x^2 - 4}}$$
### 1e) **Limit Expression:**


$$\lim_{{x \to 2}} \frac{{x - 2}}{{x^2 - 4}}$$

<br>

To solve this limit, we need to factor the denominator and simplify the expression. The denominator ( x^2 - 4 ) can be factored into ( (x + 2)(x - 2) ), which allows us to cancel out the ( x - 2 ) term in the numerator:

Expand All @@ -173,13 +134,17 @@ $$\frac{1}{4}$$

<br>

Result: The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$
[Final Result:]()

The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$


#


### 1g) $$\(\lim_{{x \to 3}} \frac{{x^3 - 27}}{{x^2 - 5x + 6}}\)$$
### 1f) **Limit Expression:**


$$\(\lim_{{x \to 3}} \frac{{x^3 - 27}}{{x^2 - 5x + 6}}\)$$

<br>

Expand All @@ -197,12 +162,18 @@ $$\

<br>

Result: The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$
[FINAL Result:]()

The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$


#

### 1h: $$\(\lim_{{x \to \infty}} \frac{{x^2}}{{2x^2 - x}}\)$$
### 1g: **Limit Expression:**


$$\(\lim_{{x \to \infty}} \frac{{x^2}}{{2x^2 - x}}\)$$

<br>

In this case, we can use L'Hôpital's rule, as the limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) when \(x\) tends to infinity.

Expand All @@ -216,16 +187,21 @@ $$\
\end{align*}
\$$

<br>

Result: The limit of the expression is $$\frac{1}{2}$$
[Final Result:]()

The limit of the expression is $$\frac{1}{2}$$

<br><br>
#

## [2. Solve the Limits:]()

## 2. Solve the Limits:

### 2a): **Limit Expression:**


### 2a): ( $\lim_{x \to \infty} \frac{1}{x^2}$ )
$\lim_{x \to \infty} \frac{1}{x^2}$

<BR>

The limit as ( x ) approaches infinity for ( $\frac{1}{{x^2}}$ ):

Expand All @@ -236,7 +212,11 @@ As ( x ) increases without bound, the value of ( \frac{1}{{x^2}} ) approaches 0
#


### 2b): ( $\lim_{x \to -\infty} \frac{1}{x^2}$ )
### 2b) **Limit Expression:**


( $\lim_{x \to -\infty} \frac{1}{x^2}$ )

<br>

The limit as ( x ) approaches negative infinity for ( $\frac{1}{{x^2}}$ ) is:


Expand All @@ -246,30 +226,25 @@ As ( x ) decreases without bound, the value of ( $\frac{1}{{x^2}}$ ) approaches

#

### 2c): ( $\lim_{x \to \infty} x^4$
)
### 2c) **Limit Expression:**


The limit as ( x ) approaches infinity for ( x^4 ) is:
grows at an increasing rate and approaches infinity for ( x^4 ) is:
$\lim_{x \to \infty} x^4$


$\lim_{{x \to \infty}} x^4 = \infty$
<br>

The limit as ( x ) approaches infinity for ( x^4 ) is:
grows at an increasing rate and approaches infinity for ( x^4 ) is:

If you have more expressions or need further assistance, feel free to ask!
$\lim_{{x \to \infty}} x^4 = \infty$

Similar to the previous expressions, the term ( 2x^5 ) grows at a faster rate than the others, causing the expression to approach infinity.

#

### 2d): ( $\lim_{{x \to \infty}} (2x^4 - 3x^3 + x + 6)$ )

The limit as ( x ) approaches infinity for ( $2x^4 - 3x^3 + x + 6$ ) is:

$\lim_{x \to \infty} (2x^4 - 3x^3 + x + 6) = \infty$

### 2d) **Limit Expression:**


As ( x ) grows larger, the term ( 2x^4 ) dominates, leading the expression to increase without bound.
$\lim_{{x \to \infty}} (2x^4 - 3x^3 + x + 6)$

#

### 2e):
<br>

The limit as ( x ) approaches negative infinity for ( 2x^4 - 3x^3 + x + 6 ) is:

Expand All @@ -280,19 +255,27 @@ Even though ( x ) is negative, the highest power term ( x^4 ) will still lead t

#

### 2f) : The limit as ( x ) approaches infinity for ( 2x^5 - 3x^2 + 6 ) is:

### 2e) **Limit Expression:**


2x^5 - 3x^2 + 6

<br>

The limit as ( x ) approaches infinity for ( 2x^5 - 3x^2 + 6 ) is:


The limit as ( x ) approaches negative infinity for ( 2x^4 - 3x^3 + x + 6 ) is:


$\lim_{x \to -\infty} (2x^4 - 3x^3 + x + 6) = \infty

Even though ( x ) is negative, y.
--
$\lim_{x \to -\infty} (2x^4 - 3x^3 + x + 6) = \infty$

#
Even though ( x ) is negative, y.

These processes demonstrates how limits help us understand the behavior of functions near points that might not be defined, by finding equivalent expressions that are easier to evaluate.

#

## These processes above demonstrates how limits help us understand the behavior of functions near points that might not be defined, by finding equivalent expressions that are easier to evaluate.

<br>

###### <p align="center"> [Copyright 2024 Quantum Software Development. Code released under the MIT License license.](https://github.com/Quantum-Software-Development/Math/blob/3bf8270ca09d3848f2bf22f9ac89368e52a2fb66/LICENSE)


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