In this post, I show that $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x \rightarrow a}g(x)}$ given that $\lim_{x\rightarrow a}f(x)=A$, $\lim_{x \rightarrow a} g(x) = B$, $B \ne 0$ and $g(x) \ne 0$. To do this, I approximately follow the steps in reference [1]. Known: From the the definition of a limit, Whenever $ 0 < |x-a| < \delta $, $ |f(x) – A| < \epsilon_1$ with $\epsilon_1 > 0$. Whenever $ 0 < |x-a| < \delta $, $ |g(x) – B| < \epsilon_2$ with $\epsilon_2 > 0$. Objective: The objective is to directly show that $ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x […]

# Category: Limit

Now that several limit properties have been proven, it is possible for me to evaluate $ \lim_{\alpha \rightarrow 0} \frac{1 – \cos \alpha}{\alpha} $. To do this, I follow the steps in Reference [1]. However, I am going to constrain $\alpha$, in radians, to be greater than or equal to zero, so that I do not need to deal with the issue of negative angles and how they are used in trigonometric identities proven with right triangles and nonnegative angles. I know that the limit definition requires negative and positive numbers if $x \rightarrow 0$, but I am going to

In this post, I show that $\lim_{x\rightarrow a}[f(x)g(x)] = \lim_{x\rightarrow a} f(x) \lim_{x\rightarrow a} g(x)$ given that $\lim_{x\rightarrow a} f(x) = A$ and $\lim_{x\rightarrow a} g(x) = B$. To do this, I approximately follow the steps in reference [1]. Known: Using the definition of a limit, $|f(x) – A|<\epsilon_1$ whenever $ 0 < |x-a| < \delta$, with $\epsilon_1 > 0$. $|g(x)- B|<\epsilon_2$ whenever $ 0 < |x-a| < \delta$, with $\epsilon_2 > 0$. Objective: The objective is to directly show that $\lim_{x\rightarrow a}[f(x)g(x)] = \lim_{x\rightarrow a} f(x) \lim_{x\rightarrow a} g(x)$. Proof: Using algebra, $ [f(x) – A] [g(x) – B] = f(x)g(x)

In this post, I show that $\lim_{x\rightarrow a}[f(x) – g(x)] = \lim_{x \rightarrow a} f(x) – \lim_{x \rightarrow a} g(x) $ given that $\lim_{x \rightarrow a} f(x) = A$ and $\lim_{x \rightarrow a} g(x) = B$. To do this, I approximately follow the steps in Reference [1]. Objective: The objective is to directly show that $\lim_{x\rightarrow a}[f(x) – g(x)] = \lim_{x \rightarrow a} f(x) – \lim_{x \rightarrow a} g(x) $. Proof: Start with the left side, $\lim_{x\rightarrow a}[f(x) – g(x)]$ From arithmetic, $=\lim_{x\rightarrow a}[f(x) + (-1)g(x)]$ From this property, $=\lim_{x\rightarrow a}f(x) + \lim_{x\rightarrow a}(-1)g(x)$ From this property, $=\lim_{x\rightarrow a}f(x) + (-1)\lim_{x\rightarrow

In this post, I show that $ \lim_{x \rightarrow a} [f(x) + g(x)]$ is equal to $\lim_{x \rightarrow a}f(x) + \lim_{x \rightarrow a} g(x)$ given that $ \lim_{x \rightarrow a}f(x) = A$ and $ \lim_{x \rightarrow a}g(x) = B$. To do this, I approximately follow the steps in Reference [1]. Objective: Using the definition of a limit, the objective is to show that: For each positive number $\epsilon_2$ there is another, $\delta$, such that whenever $ 0 < |x-a| < \delta $ we have $ |f(x)+g(x) – (A+B)| < \epsilon_2$. Proof: From the definition of a limit, it is known that For each positive number

In this post, I show that $ \lim_{x \rightarrow a} [c f(x)] = c \lim_{x \rightarrow a} [f(x)]$ if $c$ is a variable for any real number. To do this, I approximately follow the steps in reference [1]. Essential Background Information: Definition of a Limit: A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and only if, for each positive number $\epsilon$ there is another, $\delta$, such that whenever $0 < |x-a| < \delta $ we have $|f(x) – A| < \epsilon$. That is, when $x$ is near $a$ (within a distance $\delta$ from it), $f(x)$ is near

In this post, I prove that $ \displaystyle \lim_{x \rightarrow a} c = c $ if $c$ is a variable for any real number. To do this, I approximately follow the outline from reference [1]. Essential Background Information: Definition of a Limit: A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and only if, for each positive number $\epsilon$ there is another, $\delta$, such that whenever $0 < |x-a| < \delta $ we have $|f(x) – A| < \epsilon$. That is, when $x$ is near $a$ (within a distance $\delta$ from it), $f(x)$ is near $A$ (within