Dear all,(adsbygoogle = window.adsbygoogle || []).push({});

I have a question concerning calculating the following limit:

[tex]

\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\sin{(x)}}{x} = 1

[/tex]

Obviously, x=0 is not part of the domain of the function. One way to calculate the limit is using l'Hospital. Another way for these kinds of limits is using Taylor expansions (a comparable case is e.g. given in Stewart, section 11.10 example 12). The reasoning is then as follows:

(1) Write down the Taylor expansion of the numerator:

[tex]

sin{(x)} = x - \frac{x^3}{6} + \ldots

[/tex]

(2) Calculate

[tex]

\frac{\sin{(x)}}{x} = 1 - \frac{x^2}{6} + \ldots (*)

[/tex]

(3) Observe that the limit ##x\rightarrow 0## of this expression equals 1.

By expanding the numerator of ##f(x)## and dividing it by x has added the point x=0 to our original domain of ##f(x)##. So this way of calculating the limit feels a bit like cheating; we wouldn't be able to Taylor-expand ##\frac{\sin{(x)}}{x}## around x=0 because it is not part of the domain. So the Taylor expansion (*) is the Taylor expansion of our original ##f(x)## for ##x \neq 0## PLUS ##f(0)=1##. With this we have made the function continuous in ##x=0## and the limit equals the function value.

I guess this basically is the same as comparing a function ##g(x)## which is defined for all ##x##, with ##g(x)= 1 \times g(x) = \frac{x-a}{x-a}g(x)##, which is not defined for ##x=a##. Any comments?

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# I Taylor expansions, limits and domains

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