Question

Limits question :

Answer 1

@phi

Answer 2

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

Answer 3

Now i got the second in my exam limit tsin1/t which is obviously not 0/0

Answer 4

My teacher told me you can assume t= 1/x and it will be sinx/x which makes me insane.

Answer 5

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

Answer 6

but mustn't it be 0/0 in any case to apply this

Answer 7

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

Answer 8

Lol , I never thought sinx/ x was proved with a triangle tanx = sinx/x LAWL !

Answer 9

my mistake t -> infinite

Answer 10

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

Answer 11

The series definition is out of my curriculum :o

Answer 12

You know what was wrong ? it was written as 1/t -> 0 and I dumbly turned it into t-> 0 so it was meaningless .

Answer 13

Thanks phi.