Merge branch 'main' into dependabot/github_actions/github/codeql-action-3.25.7

Signed-off-by: Fabiana Campanari <fabicampanari@gmail.com>

Author: FabianaCampanari

README.md

@@ -5,9 +5,9 @@
 
 <br><br>
 
-## <p align="center"> ✍️  Resolution of Mathematics Exercises - Calculus - Limits and Derivatives
+## <p align="center"> ✍️  Resolution of Mathematics Exercises - Calculus I - Limits and Derivatives
 
-#### <p align="center"> AI Data Science - PUCSP University Repository - [Professor Eric Bacconi Gonçalves](https://www.linkedin.com/in/eric-bacconi-423137/)
+#### <p align="center"> AI Data Science - PUCSP University Math Repository - [Professor Eric Bacconi Gonçalves](https://www.linkedin.com/in/eric-bacconi-423137/)
 
 <br><br>
 
@@ -17,6 +17,9 @@
 
 ### 1a) Limit Expression:
$\lim_{{x \to 3}} \frac{{x^2 - 9}}{{x - 3}}$
 
+#### 1a)  **Limit Expression:**
+
+
 
 <!--
  <p align="center">  To solve this limit, we can use factorization:
@@ -26,15 +29,37 @@ $$\\begin{align*}
 &= \lim_{x \to 3}(x+3) \\
 &= 3+3 
 &= 6
+\large \lim_{{x \to 3}} \frac{{x^2 - 9}}{{x - 3}}
 \end{align*}
 \$$
 
 -->
 
 <br>
 
- Result: The limit of the expression is 6.
+*  **Simplified Form:** The numerator  $\large x^2 - 9$, can be factored as ( (x + 3)(x - 3) ), which simplifies the expression to:
+
+<br>
+
+$$\\begin{align*}
+\large \lim_{{x \to 3}} (x + 3)

+\end{align*}
+\$$
+
+<br>
+
+*  **Final Result:** Substituting ( x ) with 3, we get:
 
+<br>
+
+$$\large  3 + 3 = 6$$
+
+<br>
+
+*  **Explanation:** The limit as ( x ) approaches 3 for the function $\large \frac{{x^2 - 9}}{{x - 3}}$ is 6.
+
+  This is because the factor ( x - 3 ) in the denominator cancels out with the same factor in the numerator, leaving ( x + 3 ) which evaluates to 6 when ( x ) is 3.
+ 
 #
 
 ### 1b) $$\(\lim_{x \to -7}\frac{49-x^{2}}{7+x}\)$$
@@ -81,9 +106,7 @@ Therefore, the limit of the function as ( x ) approaches 1 is -2.
 
 $$f(1) = -2$$
 
-&= \frac{1}{2}
-\end{align*}
-\]
+<br>
 
 Result: The limit of the expression is -2.
 
@@ -130,7 +153,7 @@ Result: The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$
 
 <br>
 
- <p align="center">This limit can be solved using factorization and polynomial division: <br><br>
+This limit can be solved using factorization and polynomial division: <br><br>
 
 $$\
 \begin{align*}
@@ -167,60 +190,84 @@ $$\
 
  Result: The limit of the expression is $$\frac{1}{2}$$
 
+<br><br>
+
+## 2. Solve the Limits:
+
+
+### 2a): ( $\lim_{x \to \infty} \frac{1}{x^2}$ )
+
+The limit as ( x ) approaches infinity for ( $\frac{1}{{x^2}}$ ):
+
+is:
$\lim_{{x \to \infty}} \frac{1}{{x^2}} = 0$

+
+As ( x ) increases without bound, the value of ( \frac{1}{{x^2}} ) approaches 0 because the denominator grows much faster than the numerator.
+
 #
 
- <br><br>
 
+### 2b): ( $\lim_{x \to -\infty} \frac{1}{x^2}$ )
 
+The limit as ( x ) approaches negative infinity for ( $\frac{1}{{x^2}}$ ) is:

 
-<br>
+$\lim_{{x \to -\infty}} \frac{1}{{x^2}} = 0$
 
- $$x→∞lim (2x4−3x3+x+6)=∞$$
+As ( x ) decreases without bound, the value of ( $\frac{1}{{x^2}}$ ) approaches 0, similar to part a), because squaring a negative number results in a positive number, which grows larger.
 
-<br><br>
+#
 
-## 2 Limits Solutions:
+### 2c): (  $\lim_{x \to \infty} x^4$
)
 
-### 2a) 
+The limit as ( x ) approaches infinity for ( x^4 ) is:
grows at an increasing rate and approaches infinity for ( x^4 ) is:
 
-$$\lim_{x \to \infty} \frac{1}{x^2}$$
+$\lim_{{x \to \infty}} x^4 = \infty$
 
-<br>
 
-#### As x approaches infinity, the denominator ($x^2$) grows infinitely large, while the numerator remains constant at 1. This means the fraction becomes increasingly small, approaching zero.
+If you have more expressions or need further assistance, feel free to ask!
 
-<br>
+Similar to the previous expressions, the term ( 2x^5 ) grows at a faster rate than the others, causing the expression to approach infinity.
 
-Solution: 
+#
 
-$$\lim_{x \to \infty} \frac{1}{x^2} = 0$$
+### 2d): ( $\lim_{{x \to \infty}} (2x^4 - 3x^3 + x + 6)$ )
 
-<br><br>
+The limit as ( x ) approaches infinity for ( $2x^4 - 3x^3 + x + 6$ ) is:
 
-## 2b) 
+$\lim_{x \to \infty} (2x^4 - 3x^3 + x + 6) = \infty$

 
-$$\lim_{x \to -\infty} \frac{1}{x^2}$$
+As ( x ) grows larger, the term ( 2x^4 ) dominates, leading the expression to increase without bound.
 
-Similar to the previous case, as x approaches negative infinity, the denominator ($x^2$) grows infinitely large (but remains positive), and the numerator remains constant at 1. 
+#
 
-<BR>
+### 2e):
 
-The fraction becomes increasingly small, approaching zero.
+The limit as ( x ) approaches negative infinity for ( 2x^4 - 3x^3 + x + 6 ) is:
 
-<BR>
+$\lim_{x \to -\infty} (2x^4 - 3x^3 + x + 6) = \infty$

 
-Solution: 
+Even though ( x ) is negative, the highest power term ( x^4 )  will still lead the expression to increase without bound because the even power makes it positive.
 
-$$\lim_{x \to -\infty} \frac{1}{x^2} = 0$$
 
 #
 
-3c) 
+### 2f) : 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.
+
+
+
+  
+
 
 
 
 
 
+ 
+
+
 
 
 
@@ -247,6 +294,9 @@ $$\lim_{x \to -\infty} \frac{1}{x^2} = 0$$
 
 
 
+
+
+<!--
  <br><br>
 
  ## 3.Calculate the Following Limits
@@ -287,44 +337,6 @@ $$f(x)=cxm+fxm−1+...+gx+haxn+bxn−1+...+dx+e$$
 In this case, the highest-order term in the numerator is 2x4, and the highest-order term in the denominator is x3. 
 
 As x approaches infinity, 2x4 grows much faster than x3, and so the function f(x) approaches zero.
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