limits

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Reply 1

topcrabcakes

It tends to infinity.

Reply 2

topcrabcakes

Sorry im being silly, do u mean what happens to the limit as x tends to infinity? If so, sin x/x tends to 0.

Reply 3

Manifest

wacabac

Sorry im being silly, do u mean what happens to the limit as x tends to infinity? If so, sin x/x tends to 0.

As x tends to infinity

for +ve infinity, sinx/x tends to -ve infinity
for -ve infinity, sinx/x tends to +ve infinity

Manifest

Reply 4

topcrabcakes

Manifest

As x tends to infinity

for +ve infinity, sinx/x tends to -ve infinity
for -ve infinity, sinx/x tends to +ve infinity

Manifest

How does that work? sin x is periodic and 1/x --> 0 as x--->infinity

Reply 5

steve2005

Manifest

As x tends to infinity

for +ve infinity, sinx/x tends to -ve infinity
for -ve infinity, sinx/x tends to +ve infinity

Manifest

I disagree. As x tends to infinity sinx/x tends to zero.

The maximum value of sinx is one. As the denominator gets larger the quotient gets smaller. When x is very large the quotient is very very small.... ie tends to zero. ( At least that's what I think)

Edit:
I see Manifest is at Oxford... so I hope he has made a mistake... I'm a little worried I might be making a fool of myself.

Manifest is in 6th form... sorry misread details.

Reply 6

Infinity_Kev

its (between 1 and 0) over infinity

anything over infinity will be zero right?

Reply 7

Gaz031

Zagani

what happens when the limit of sinx/x tends to infinity?

It tends to zero by the sandwich rule, since we have -1/x<=sinx/x<=1/x and (-1/x)->0 and (1/x)->0.

Reply 8

Manifest

steve2005

First of all, I am not in oxford. I made a silly mistake by giving tanx properties to sinx. (both of them are odd functions anyway.)

Reply 9

Dekota

Zagani

what happens when the limit of sinx/x tends to infinity?

I think you mean:

'Limit of (sinθ )/θ, as θ -> 0"

The limit is 1.

Reply 10

Infinity_Kev

Just out of intrest what is the limit of (Tan@)/@ as @ tends to ∞? And an explanation would be cool :tsr:

Reply 11

Dekota

Infinity_Kev

Just out of intrest what is the limit of (Tan@)/@ as @ tends to ∞? And an explanation would be cool :tsr:

(Tan@)/@ = (sin@/cos@)/@
..............= lim[1/@].(sin@/cos@)
..............= lim[1/@] = 0

Hence (Tan@)/@ -> 0, as @ -> ∞

Reply 12

Infinity_Kev

Is converting Tans to Sins and Coss a standard technique for limits of trig functions?

Reply 13

Dekota

Infinity_Kev

Is converting Tans to Sins and Coss a standard technique for limits of trig functions?

No; I was just about to mention that, my method wasn't the 'standard', but then again I haven't studied limits ... so :p:

Can someone provide a better explanation of the limit (tan@)/@ ? ... :redface:

Reply 14

AlphaNumeric

V.P. Keys

Hence (Tan@)/@ -> 0, as @ -> ∞

sin and cos are always between -1 and 1, but tan is between -infinity and infinity, so your assumption that sin(x)/cos(x) -> 1 as x->infinity is wrong. I suspect it has no limit.

Reply 15

Dekota

V.P. Keys

(Tan@)/@ = (sin@/cos@)/@
..............= lim[1/@].(sin@/cos@)
..............= lim[1/@] = 0

Hence (Tan@)/@ -> 0, as @ -> ∞

Actually looking at this, you can just do:

lim[tan@/@] = lim[tan@].lim[1/@]
.................=> lim[1/@] -> 0
∴ (tan@)/@ -> 0, as @ -> ∞

similarly, (sin@)/@ -> 0, as @ -> ∞
............(cos@)/@ -> 0, as @ -> ∞

Reply 16

Manifest

where is alpha numeric?

Reply 17

Gaz031

V.P. Keys

Surely the bolded part assumes that tan@ is eventually bounded, which it isn't.

Reply 18

Dekota

Gaz, can you then please provide a better solution/explanation.

Haven't done limits you see. :P

Reply 19

dvs

tanx/x = sinx/x . 1/cosx

We know that sinx/x -> 0, so we have to check how 1/cosx behaves. As it turns

\lim_{x \rightarrow \infty} \frac{1}{\cos x}

doesn't exist. (To prove this consider the two sequences x_n = (2n)pi and y_n = (2n+1)pi and note that they increase beyond all bounds, whereas 1/cos(x_n) -> 1 and 1/cos(y_n) -> -1.) Hence