How do you find the limit of #2x-1# as #x->0#?

1 Answer
Jim H
Dec 31, 2016

See below.

Explanation:

If #x# is very, very close to #0#, then #2x# is close to "#2" times " 0#", which is #0#.

And #2x-1# or "#2x" minus "1#", which is close to #0-1# which is #-1#.

So,

#lim_(xrarr0) (2x-1) = -1#.

Note 1: We don't usually write out the reasoning I presented.

Note 2: The function #f(x) = 2x-1# is well-behaved in the sense that we can find the limit at any value of #x# by plugging in the number (by substitution).

We call functions that are well behaved in this way "continuous" functions.

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