limits (1 Viewer)

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hmm..this was in my assignment and i couldn't do it, and later the uni uploaded the solutions

Q:
evaluate
lim_x-->infinity (sin2x) /x

A:
Observe that -1 <= sin x <= 1 for all x. For x = 0, we have
-1/x <= sin x /x <= 1/x
since both 1/x and -1/x tend to 0 as x --> infinity, we obtain:
lim_x-->infinity (sin2x) /x = 0


umm..could someone briefly explain why
like the +- 1/x part

thanks

 

[Xayma]

which bit the inequality or the limits of those?

The first inequality is just putting everything on x

Eg 1<2
Therefore 1/3<2/3
Etc

The second one is as x gets very large 1/x approaches 0 (from the positive side)

As x gets very large -1/x approaches 0 (from the negative side)

 

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oh i think i get it now, they r only trying to state the obivious
-__________-;;

ok, heres another one

lim_{x --> 0} x² sqrt[1 + cos(1/x)]

the answer reads
by using 0 <= sqrt[1 + cos(1/x)] <= sqrt(2) and using squeezing principle, we can prove it..



lol

 

[turtle_2468]

Therefore
0 <= exp <=x^2*sqrt(2).
LHS obviously const
RHS obviously -> 0 as x -> 0
Hence by squeezing principle (inequalities on both sides but going to same thing as limits are taken) the middle thing is that limit as well ie 0.

 

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Originally posted by turtle_2468
Therefore
0 <= exp <=x^2*sqrt(2).

hmm..where did x² come from

and i was wondering..
why is it that its sqrt(2)??? not 1

 

[CM_Tutor]

Maximum possible value of cos(1 / x) is 1, and minimum value is -1.
ie. -1 <= cos(1 / x) <= 1
0 <= 1 + cos(1 / x) <= 2, on adding 1 to LHS, MHS and RHS
0 <= sqrt(1 + cos(1 / x) <= sqrt(2), on sqaure rooting
0 <= x² * sqrt(1 + cos(1 / x)) <= x² * sqrt(2), on multiplying by x², which we do to get the desired expression.

Now, taking limits as x --> 0, we get: 0 <= lim (x->0) [x² * sqrt(1 + cos(1 / x))] <= 0
and so lim (x->0) [x² * sqrt(1 + cos(1 / x))] = 0

 

[redslert]

hmm would you normally test for
limit -> +0
limit -> -0

i know for this question doesn't really matter as it is both a square and cos is always + for all x
but if it was another question would you test +/-0 or +/- infinity or +/-whatever