From 8f9f8a57af512ad6a29118270fd674c6c97e45ce Mon Sep 17 00:00:00 2001 From: Fabiana Campanari Date: Tue, 13 Aug 2024 17:09:06 -0300 Subject: [PATCH] Update README.md Signed-off-by: Fabiana Campanari --- README.md | 30 ++++++++++++------------------ 1 file changed, 12 insertions(+), 18 deletions(-) diff --git a/README.md b/README.md index 8bb9c7a..2dc3b25 100644 --- a/README.md +++ b/README.md @@ -33,7 +33,7 @@ $$lim_{{x \to 3}} \frac{{x^2 - 9}}{{x - 3}}$$
-* **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:
@@ -44,7 +44,7 @@ $$\\begin{align*}
-* [**Final Result:**]() +[**Final Result:**]() Substituting ( x ) with 3, we get: @@ -52,9 +52,9 @@ $$\large 3 + 3 = 6$$
-* **Explanation:** The limit as ( x ) approaches 3 for the function $\large \frac{{x^2 - 9}}{{x - 3}}$ is 6. +**Explanation:** The limit as ( x ) approaches 3 for the function $\large \frac{{x^2 - 9}}{{x - 3}}$ is 6. - This is because the factor ( x - 3 ) in the denominator cancels out with the same factor in the numerator, leaving ( x + 3 ) which evaluates to 6 when ( x ) is 3. +This is because the factor ( x - 3 ) in the denominator cancels out with the same factor in the numerator, leaving ( x + 3 ) which evaluates to 6 when ( x ) is 3. # @@ -64,25 +64,25 @@ $$\lim_{{x \to -7}} \frac{{49 - x^2}}{{7 + x}}$$
-* **Simplified Form:** The numerator $\large ( 49 - x^2 )$ is a difference of squares and can be factored as $\large (7 + x)(7 - x)$. +**Simplified Form:** The numerator $\large ( 49 - x^2 )$ is a difference of squares and can be factored as $\large (7 + x)(7 - x)$.
-* **This allows us to simplify the expression by canceling out the common factor of:** $\large ( 7 + x )$ in the numerator and denominator: +**This allows us to simplify the expression by canceling out the common factor of:** $\large ( 7 + x )$ in the numerator and denominator: $$\large \lim_{{x \to -7}} (7 - x)
$$
-* [**Final Result:**]() +[**Final Result:**]() When we substitute ( x ) with -7, the expression simplifies to:
7 - (-7) = 14
-* **Explanation:** The limit of the function $(\Large \frac{{49 - x^2}}{{7 + x}} )$ as ( x ) approaches -7 is 14. - - This result is obtained because after canceling the common factor, we are left with ( 7 - x ), which equals 14 when ( x ) is -7. +**Explanation:** The limit of the function $(\Large \frac{{49 - x^2}}{{7 + x}} )$ as ( x ) approaches -7 is 14. + +This result is obtained because after canceling the common factor, we are left with ( 7 - x ), which equals 14 when ( x ) is -7. # @@ -137,10 +137,8 @@ $$\frac{1}{4}$$ The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$ - # - ### 1f) **Limit Expression:**
 $$\(\lim_{{x \to 3}} \frac{{x^3 - 27}}{{x^2 - 5x + 6}}\)$$ @@ -165,7 +163,6 @@ $$\ The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$ - # ### 1g: **Limit Expression:**
 @@ -186,10 +183,9 @@ $$\ \end{align*} \$$ - [Final Result:]() - - The limit of the expression is $$\frac{1}{2}$$ + +The limit of the expression is $$\frac{1}{2}$$ # @@ -210,7 +206,6 @@ As ( x ) increases without bound, the value of ( \frac{1}{{x^2}} ) approaches 0 # - ### 2b) **Limit Expression:**
 ( $\lim_{x \to -\infty} \frac{1}{x^2}$ ) @@ -251,7 +246,6 @@ $\lim_{x \to -\infty} (2x^4 - 3x^3 + x + 6) = \infty$
 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. - # ### 2e) **Limit Expression:**