What is $ \lim_{x\rightarrow0} \frac{\sin x}{x}$ ? Recall the definition of a limit, repeated here for reference [2]. 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 a distance $\epsilon$ from it). In symbols we write $\lim_{x \rightarrow a} f(x) = A$. Using this definition, $a$ is $0$ in this case. Now I extract as much information from the limit definition as possible, for this case of $a$ of $0$. Using the language from reference [2], and replacing $a$ with $0$, one has: A function $f(x)$ approaches a limit $A$ as $x$ approaches $0$ if, and only if, for each positive number $\epsilon$ there is another, $\delta$, such that whenever $0 < |x| < \delta $ we have $|f(x) – A| < \epsilon$. That is, when $x$ is near $0$ (within a distance $\delta$ from it), $f(x)$ is near $A$ (within a distance $\epsilon$ from it). In symbols we write $\lim_{x \rightarrow 0} f(x) = A$. Using the previous post about the definition of a limit, it is possible to obtain information about $x$ and $f(x)$ for the case of $a=0$. It is helpful to draw the situation. This analysis applies for each positive number that $\delta$ can be, which implies that $\delta$ can be very close to zero without being zero. To find the limit, one must find the value of $A$. How should this be done? In this case, $f(x)=\frac{\sin x}{x}$. Perhaps it possible to view this situation graphically, by using the values of sine with angles in radians. Here, the […]
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