Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/46

22 If now $$x$$ approaches the limit zero,$$\underset{x=0}{\operatorname{limit}}\;\frac{x}{\sin x}$$must lie between the constant $$1$$ and $$\underset{x=0}{\operatorname{limit}}\frac{1}{\cos x}$$, which is also $$1$$.

It is interesting to note the behavior of this function from its graph, the locus of equation$$y=\frac{\sin x}{x}.$$

Although the function is not defined for $$x=0$$, yet it is not discontinuous when $$x=0$$ if we define

One of the most important limits in the Calculus is$$\underset{x=0}{\operatorname{limit}}\;(1+x)^{\frac{1}{x}}=2.71828\cdots=e.$$

To prove rigorously that such a limit $$e$$ exists, is beyond the scope of this book. For the present we shall content ourselves by plotting the locus of the equation$$y=(1+x)^{\frac{1}{x}}$$and show graphically that, as $$x\doteq 0$$, the function $$(1+x)^{\frac{1}{x}}(=y)$$ takes on values in the near neighborhood of $$2.718\cdots$$, and therefore $$e=2.718\cdots$$ approximately.