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MIT 18.01 Single Variable Calculus (OCW) 20 Online
OpenStudy (anonymous):

Basic question: In the 2nd lecture on the use of limits for proving continuity of differentiable functions, is lim { [f(x) - f(x0)]/[x-x0] } = 1/[x-x0] lim [f(x) - f(x0)] if we see lim as an operator?

OpenStudy (waynex):

\[\lim_{x \rightarrow x_{0}} \left( \frac{ f(x)-f(x_{0}) }{ x-x_{0} } \right) = \frac{1}{x-x_{0}} \lim_{x \rightarrow x_{0}} \left( f(x)-f(x_{0}) \right)\] If you are asking if the right hand side represents a valid algebraic manipulation of the left hand side, the answer is no. In the denominator, x is not a constant; therefore, it cannot be moved out of the limit.

OpenStudy (anonymous):

no you can not consider it as an operator. however, it is more beneficial to [/delta x/] instead of \[x-x_0\]

OpenStudy (anonymous):

x -> x0, then f(x) - f(x0) is changing but also x - x0 is changing therefore you cannot treat x-x0 as constant and put it before limit

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