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limit sin x/ x x-> 0 this one is 0/0 but assume x = 1 /t limit tsin1/t 1/t -> 0 so t->0
Now i got the second in my exam limit tsin1/t which is obviously not 0/0
My teacher told me you can assume t= 1/x and it will be sinx/x which makes me insane.
limit as x->0 of (sin x) / x is a famous limit The derivation of this limit that I have seen rely on geometric arguments https://www.khanacademy.org/math/differential-calculus/limits_topic/squeeze_theorem/v/proof-lim-sin-x-x
but mustn't it be 0/0 in any case to apply this
I assume you mean the problem is limit t sin1/t as t->infinity rewrite as x=1/t , x-> 0 and the problem is limit sin(x)/x as x->0 which = 1
Lol , I never thought sinx/ x was proved with a triangle tanx = sinx/x LAWL !
my mistake t -> infinite
The other way to find the limit of sin x / x is to use the series definition of sin x https://en.wikipedia.org/wiki/Trigonometric_functions#Series_definitions sin x = x - x^3/3! + x^5/5! - ... now divide by x sin x / x = 1 - x^2/3! + x^4/5! - .... now let x->0 to get limit sin x / x = 1
The series definition is out of my curriculum :o
You know what was wrong ? it was written as 1/t -> 0 and I dumbly turned it into t-> 0 so it was meaningless .