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Two Questions About MATHEMATICS : LIMIT?
Please prove that :
1) lim sin x equal as lim x = 1
x--> 0 x x--> 0 sin x
(note : between sin x and x or x and sin x there are lines :) it's a fraction )
2) lim tan x equal as lim x = 1
x--> 0 x x--> 0 tan x
(note : between tan x and x or x and tan x there are lines :) it's a fraction )
4 Answers
- Anonymous1 decade agoFavorite Answer
You may use l'Hospital rule to compute those limits.
If lim f(x)/g(x) when x->a leads to an indermination like 0/0, then
lim f(x)/g(x) = lim (df/dx)/(dg/dx) when x->a
1) lim (sin x)/x = lim (cos x)/1 = 1
lim x/(sin x) = lim 1/(cos x) = 1
2) lim (tan x)/x = lim (1/(cos x)^2)/1 = 1
lim x/(tan x) = lim 1/(1/(cos x)^2) = 1
- 1 decade ago
To do both of these limits you must use L'Hôpital's rule, which says that when you plug in the limit value, if it is unsolvable such as the case of something divided by zero, you can take the derivative of both the top and bottom and take the limit again. Hopefully I can clear this up by working the problems.
1) lim(x->0) sin(x)/x If you plug in x=0, you get sin(0)/0 which is unsolvable so we take the derivative of the top and bottom separately, and take the limit again.
lim(x->0) cos(x)/1 Plugging in x=0 we get cos(0)/1 = 1/1 = 1
This will be the same if it is flipped, and also the same for problem number 2.
I dont feel like this is very clear, but maybe it helped.
- Anonymous5 years ago
the last one
