Asked by Michael
We are doing limits in Calculus, but now we are doing trig limits too and i do not get how to do any of them.
For example ones like this, how do you do them?
sec (x-1) / (x sec x) x approaches 0
((3 sin (x)) 1 - cos(x)) / (x^2) x approaches 0
(sin (2x)) / (sin (3x)) x approaches 0
(1-tan(x)) / (sin (x) - cos (x)) x approaches 0
I just want to learn how to these kinds of limits so i know how to do all the types.
Please help, is there a website that will teach me how to do these kinds of limits?
Thank You.
For example ones like this, how do you do them?
sec (x-1) / (x sec x) x approaches 0
((3 sin (x)) 1 - cos(x)) / (x^2) x approaches 0
(sin (2x)) / (sin (3x)) x approaches 0
(1-tan(x)) / (sin (x) - cos (x)) x approaches 0
I just want to learn how to these kinds of limits so i know how to do all the types.
Please help, is there a website that will teach me how to do these kinds of limits?
Thank You.
Answers
Answered by
drwls
If going to the limit results in a 0/0 or infinity/innfinity indeterminate fraction, then take the ratio of the derivatives of the numerator and denominator to get the limit. That is called L'Hopital's rule.
For the question
Lim (sin (2x)) / (sin (3x))
x->0
the answer is [2 (cos(0)]/[3 cos(0)] = 2/3
Remember that sin x is very close to x when x is small. this tells you right away that it approaches 2x/3x = 2/3
For a website that should help, Google "L'Hopital's rule".
One good site is
http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
For the question
Lim (sin (2x)) / (sin (3x))
x->0
the answer is [2 (cos(0)]/[3 cos(0)] = 2/3
Remember that sin x is very close to x when x is small. this tells you right away that it approaches 2x/3x = 2/3
For a website that should help, Google "L'Hopital's rule".
One good site is
http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
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