limit
prove that \[\lim_{x \rightarrow a} f(x)\] =L
if and only if \[\lim_{x \rightarrow a} (f(x)-L)\]
does: \[\lim_{x \rightarrow a}(f(x)-L) = 0\] ?
if so: \[\lim_{x \rightarrow a}(f(x)-L)=\lim_{x \rightarrow a}f(x) + \lim_{x \rightarrow a}(L)\]
by linearity
the second term is just L \[\lim_{x \rightarrow a}f(x) - L = 0\]
Sorry, three posts up I made a typo, should be limf(x) - lim(L)
anyway, then through algebra, you get: \[\lim_{x \rightarrow a}f(x) = L\]
okk
here are some limit properties: http://tutorial.math.lamar.edu/Classes/CalcI/LimitsProperties.aspx
what about \[\lim_{x \rightarrow a} |f(x)| =|L|\]
then you have: \[\pm \lim_{x \rightarrow a}f(x) = \pm L\]
Out of that, you get two cases: \[\lim_{x \rightarrow a}f(x) =L\] and \[\lim_{x \rightarrow a}f(x) =-L\]
how to prove left hand side is equal right hand side
list out the four cases
\[\lim_{x \rightarrow a}f(x) = L\] \[\lim_{x \rightarrow a}f(x) = -L\] \[-\lim_{x \rightarrow a}f(x) = L\] \[-\lim_{x \rightarrow a}f(x) = -L\]
the negatives cancel in the last one
and you can just divide by -1 in the second one
my question is prove that \[\lim_{x \rightarrow a} |f(x)| = |L|\] if and only if \[\lim_{x \rightarrow a}( f(x)-L)\]
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