diff --git a/README.md b/README.md
index d248d36..8b3671a 100644
--- a/README.md
+++ b/README.md
@@ -34,7 +34,7 @@ $$lim_{{x \to 3}} \frac{{x^2 - 9}}{{x - 3}}$$
-* **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:
@@ -45,9 +45,9 @@ $$\\begin{align*}
-* **Final Result:** Substituting ( x ) with 3, we get:
+* [**Final Result:**]()
-
+Substituting ( x ) with 3, we get:
$$\large 3 + 3 = 6$$
@@ -75,7 +75,9 @@ $$\large \lim_{{x \to -7}} (7 - x)
$$
-* **Final Result:** When we substitute ( x ) with -7, the expression simplifies to:
7 - (-7) = 14
+* [**Final Result:**]()
+
+When we substitute ( x ) with -7, the expression simplifies to:
7 - (-7) = 14
@@ -85,65 +87,18 @@ $$\large \lim_{{x \to -7}} (7 - x)
$$
#
-#### 1c) 8 88**Limit Expression:**
-
-$$lim_{{\Large \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}}$$
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+### 1c) **Limit Expression:**
+$$lim_{{\Large \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}}$$n
+
-
+### 1d) **Limit Expression:**
-#### 1e) $$\lim_{{x \to 1}} \frac{{x^2 - 2x + 1}}{{x - 1}}$$
+$$\lim_{{x \to 1}} \frac{{x^2 - 2x + 1}}{{x - 1}}$$
+
To calculate the limit, we can simplify the expression by factoring the numerator, which is a perfect square trinomial. Factoring (x^2 - 2x + 1), we get ((x - 1)(x - 1)). The denominator is already in factored form as (x - 1). Thus, the function simplifies to:
@@ -157,11 +112,17 @@ Since there are no more terms that depend on ( x ), this simplifies to:
$$\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}}$$
+
+
To solve this limit, we need to factor the denominator and simplify the expression. The denominator ( x^2 - 4 ) can be factored into ( (x + 2)(x - 2) ), which allows us to cancel out the ( x - 2 ) term in the numerator:
@@ -173,13 +134,17 @@ $$\frac{1}{4}$$
-Result: The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$
+[Final Result:]()
+
+The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$
#
-### 1g) $$\(\lim_{{x \to 3}} \frac{{x^3 - 27}}{{x^2 - 5x + 6}}\)$$
+### 1f) **Limit Expression:**
+
+$$\(\lim_{{x \to 3}} \frac{{x^3 - 27}}{{x^2 - 5x + 6}}\)$$
@@ -197,12 +162,18 @@ $$\
-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: $$\(\lim_{{x \to \infty}} \frac{{x^2}}{{2x^2 - x}}\)$$
+### 1g: **Limit Expression:**
+
+$$\(\lim_{{x \to \infty}} \frac{{x^2}}{{2x^2 - x}}\)$$
+
+
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.
@@ -216,16 +187,21 @@ $$\
\end{align*}
\$$
-
- Result: The limit of the expression is $$\frac{1}{2}$$
+[Final Result:]()
+
+ The limit of the expression is $$\frac{1}{2}$$
-
+#
+
+## [2. Solve the Limits:]()
-## 2. Solve the Limits:
+### 2a): **Limit Expression:**
-### 2a): ( $\lim_{x \to \infty} \frac{1}{x^2}$ )
+ $\lim_{x \to \infty} \frac{1}{x^2}$
+
+
The limit as ( x ) approaches infinity for ( $\frac{1}{{x^2}}$ ):
@@ -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:**
+
+( $\lim_{x \to -\infty} \frac{1}{x^2}$ )
+
+
The limit as ( x ) approaches negative infinity for ( $\frac{1}{{x^2}}$ ) is:
@@ -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:
+$\lim_{x \to \infty} x^4$
-$\lim_{{x \to \infty}} x^4 = \infty$
+
+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):
+
The limit as ( x ) approaches negative infinity for ( 2x^4 - 3x^3 + x + 6 ) is:
@@ -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
+
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+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.
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###### [Copyright 2024 Quantum Software Development. Code released under the MIT License license.](https://github.com/Quantum-Software-Development/Math/blob/3bf8270ca09d3848f2bf22f9ac89368e52a2fb66/LICENSE)