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# Limits Question watch

1. If anyone could just show me how to do this limit it would be much appreciated. My mind's just gone completely blank

2. Eek! Wouldn't know where to start!! I would, if I had to have a stab, use the MacLaurin/Taylor series, but that's probably the entirely wrong thing to do...

Just forget about me... I've only covered AS
3. (Original post by henryt)
Eek! Wouldn't know where to start!! I would, if I had to have a stab, use the MacLaurin/Taylor series, but that's probably the entirely wrong thing to do...

Just forget about me... I've only covered AS
I've tried to get Maple to do the series for x.tan(1/x) but it just says Error, so I don't think that's how to go about it annoyingly
4. The limit is the same as tan(y)/y as y goes to zero. But this is

(sin(y)/y)/cos(y)

And we know that sin(y)/y tends to 1 as y tends to zero (use the maclaurin series, although it's not entirely rigorous unless you know all the theory). Also, cos(y) tends to 1 as y tends to zero so the required limit is 1.
5. (Original post by mikesgt2)
The limit is the same as tan(y)/y as y goes to zero. But this is

(sin(y)/y)/cos(y)

And we know that sin(y)/y tends to 1 as y tends to zero (use the maclaurin series, although it's not entirely rigorous unless you know all the theory). Also, cos(y) tends to 1 as y tends to zero so the required limit is 1.
I agree with that.
6. let u = 1/x
The limit can be written as
1/u (sinu/cosu) while u tends to 0
= (sinu)/u because cos u tends to 1
= 1 well I think that's a standard result you need to learn
7. (Original post by olliemccowan)
If anyone could just show me how to do this limit it would be much appreciated. My mind's just gone completely blank

let w=1/x

we want lim w->0 tanw/w = lim->0 sec^2w =1 Using L'hopitals Rule.
8. (Original post by mikesgt2)
The limit is the same as tan(y)/y as y goes to zero. But this is

(sin(y)/y)/cos(y)

And we know that sin(y)/y tends to 1 as y tends to zero (use the maclaurin series, although it's not entirely rigorous unless you know all the theory). Also, cos(y) tends to 1 as y tends to zero so the required limit is 1.
Thanks I've just got to get into the swing of things so I can pick up on the little things like that which make the question so simple.

9. the taylor eaxpansion of tan 1/x is contains infinite terms starting
1/x + 1/3x^3 + ...
Mulitiply this by x and you get
1 + 1/3x^2 + ...
all terms other than 1 go to 0 as x goes to infinity so by Algebra of limits the sum goes to 1

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Updated: April 14, 2006
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