Limits are the devil. (1 Viewer)

[Bookie]

My MX1 test is on Thursday [greetings fellow High boys], and although most other topics are fairly do-able [even though MX1 is pretty tough], I seem to struggle with these limits. Although I think I understood them about 2 months ago when we first learnt them, I have jack all idea about such ones as the following. Please attempt.

cheers.

 

[Yip]

Im not sure, but i'll take a stab

1. lim(h->0) [sin(x+h)-sin(x-h)]/2h]
= lim(h->0) [2sin(h)cos(x)/2h]
=lim(h->0) [sinh/h].[cosx]
=1.cosx=cosx

2. lim (x->0) [xsin(pi/x)]
= lim (x->0) {([xsin(pi/x)]/[pi/x]).(pi/x)}
=lim (x->0) {([sin(pi/x)]/[pi/x]).x(pi/x)}
=lim (x->0) {([sin(pi/x)]/[pi/x]).pi}
=pie
i think ~.~

 

[Stan..]

They are givens, your meant to use them. Or not.

 

[Yip]

are u sure stan? those first 2 limits arent equal to anything, looks like an evaluate to me..........

 

[Slidey]

Dear Bookie,

For this limit, I suspect you should simply use your judgement. As x->infinity, the function has some odd behaviour. However, as x approaches 0, both the sin and the x approach zero (with more odd behaviour very close by on either side), so it makes sense tat the entire function also approaches 0.

lim xsin(pi/x)
x->0
=0

For number 1, Yip is correct (well done). For number 2 he has made a nice leap, however it is unfortunately incorrect, as sin(1/x)/1/x equals one only as x approaches infinity (not zero).

Number 3 makes absolutely no sense. For starters, a limit in x cannot possibly equal a function in x, and moreover the limit of the LHS as x-> 0 is 4, while on the RHS (if we take it) it is 1.

 

[Bookie]

Cheers Slide rule, yeh, and number 3 was a stuff up. Sorry about that.

I understand 1, I was too lazy to expand sin(x + h). Thats understandable.

Yip got 2 correct according to the book, but i want to know how he gets from;

= lim (x->0) {([xsin(pi/x)]/[pi/x]).(pi/x)}

to

=lim (x->0) {([sin(pi/x)]/[pi/x]).x(pi/x)}

. Looking back as past papers, there aren't many limits, but I'd still like to know.

EDIT - actually dont worry, I just subsituted (pi/x) with 'u'. Makes it a bit more understandable. Thanks anyway my friends.

 

[acmilan]

He took the x from in front of the sin(pi/x) out of the bracket.

 

[icycloud]

Question (2) only equals "pi" if x --> infinity. (I would venture to say it was a typo in the original post.)

And as Slide Post said, question (3) equals 4, not Cos[x]. As follows:

lim {x-->0} (3Sin[x] - Sin[3x])/(x³)
= lim {x-->0} (3Sin[x] - Sin[2x+x])/(x³)
= lim {x-->0} (3Sin[x] - Sin[2x]Cos[x] - Sin[x]Cos[2x])/(x³)
= lim {x-->0} (3Sin[x] - 2Sin[x]Cos²[x] - Sin[x](1-2Sin²[x]))/(x³)
= lim {x-->0} (3Sin[x] - 2Sin[x](1-Sin²[x]) - Sin[x] + 2Sin³[x])/(x³)
= lim {x-->0} (3Sin[x] - 2Sin[x] + 2Sin³[x] - Sin[x] + 2Sin³[x])/(x³)
= lim {x-->0} (4Sin³[x])/(x³)
= 4 * lim {x-->0} (Sin[x]/x)³
= 4 * 1³
= 4 #

 

[Riviet]

Very nice icycloud, i understood it all the way, nice job.