we know that sinx/x where lim x->0 is equal to 1 but is the reciprocal also equal to one.
2007-01-26 00:27:18 · 7 answers · asked by vishnu c 1 in Science & Mathematics ➔ Mathematics
Yes. Use L'Hospital' theorem:
lim x/sin(x), x->0 = lim x'/sin'(x) = lim 1/cos(x) = 1
2007-01-26 00:33:31 · answer #1 · answered by catarthur 6 · 1⤊ 0⤋
A way to do this is to graph the function y = (tan x - x)/(x - sin x) and see how the graph behaves as x approaches 0 from the right. If this is done you will see that the value of the function approaches 2, therefore the Lim (x -> 0+) (tan x - x)/(x - sin x) = 2
2016-03-29 03:19:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Yes, this is consequence of the fact that if lim (x -> a) f(x) = c and cis different from 0, than lim (x -> a) 1/f(x) = 1/c.
2007-01-26 00:55:12 · answer #3 · answered by Steiner 7 · 2⤊ 0⤋
As long as the limit exists as a real number c, the reciprocal limit is 1/c.
2007-01-26 00:36:46 · answer #4 · answered by Professor Maddie 4 · 1⤊ 0⤋
Supposedly yeah.
According to L'Hopital Rule;
lim f(x) = 0
lim g(x) = 0
Then lim[f(x)/g(x)] = 1
2007-01-26 00:40:47 · answer #5 · answered by Alzis 1 · 0⤊ 3⤋
yes, just look at the graph
2007-01-26 00:35:56 · answer #6 · answered by connor0314 3 · 0⤊ 0⤋
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2007-01-26 00:33:16 · answer #7 · answered by gianlino 7 · 0⤊ 1⤋