diff --git a/README.md b/README.md index 4e0a9a0..fbc4f88 100644 --- a/README.md +++ b/README.md @@ -13,7 +13,7 @@ ## 1. Find the limits: -#### 1a) $$\(\lim_{x \to 3}\frac{x^{2}-9}{x-3}\)$$ +### 1a) $$\(\lim_{x \to 3}\frac{x^{2}-9}{x-3}\)$$

To solve this limit, we can use factorization: @@ -33,7 +33,7 @@ $$\\begin{align*} # -#### 1b) $$\(\lim_{x \to -7}\frac{49-x^{2}}{7+x}\)$$ +### 1b) $$\(\lim_{x \to -7}\frac{49-x^{2}}{7+x}\)$$ Again, we can use factorization: @@ -49,7 +49,7 @@ Result: The limit of the expression is -14. # -#### 1c) $$\lim_{{x \to 0}} \frac{x^3}{2x^2 - x}$$ +### 1c) $$\lim_{{x \to 0}} \frac{x^3}{2x^2 - x}$$ We factor out ( x ) from the denominator: @@ -63,7 +63,7 @@ Result: The limit of the expression is 0. # -#### 1d) $$f(x) = \lim_{x \to 1} \frac{x^2 - 4x + 3}{x - 1}$$ +### 1d) $$f(x) = \lim_{x \to 1} \frac{x^2 - 4x + 3}{x - 1}$$ To calculate the limit, we can simplify the expression by factoring the numerator, which results in: @@ -85,7 +85,7 @@ Result: The limit of the expression is -2. # -#### 1e) $$\lim_{{x \to 1}} \frac{{x^2 - 2x + 1}}{{x - 1}}$$ +### 1e) $$\lim_{{x \to 1}} \frac{{x^2 - 2x + 1}}{{x - 1}}$$ To calculate the limit, we can simplify the expression by factoring the numerator, which is a perfect square trinomial. Factoring (x^2 - 2x + 1), we get ((x - 1)(x - 1)). The denominator is already in factored form as (x - 1). Thus, the function simplifies to: @@ -104,7 +104,7 @@ Result: The limit of the function as ( x ) approaches 1 is simply 0. # -#### 1f) $$\lim_{{x \to 2}} \frac{{x - 2}}{{x^2 - 4}}$$ +### 1f) $$\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: