uni calculus (limits) (1 Viewer)

[...]

hmm..

on a different note, say
f(x) = [sqrt(1 + 16x²) - 1] / 2x

explain why f(x) = 1 has exactly one solution in the interval (0, oo), stating clearly any results that you use

so i start from 1 = [sqrt(1 + 16x²) - 1] / 2x ??

 

[ND]

Yep start with that:

1=[sqrt(1+16x^2)-1]/2x
2x=sqrt(1+16x^2)-1
The roots of that will be the pts of intersection between the line y=2x, and the curve y=sqrt(1+16x^2)-1. Now since d/dx (2x) is greater than d/dx (sqrt(1+16x^2)-1) for a very small +ve value of x, and d/dx(d/dx (sqrt(1+16x^2)-1)) (i.e. the 2nd derivative) is +ve (and it's also obvious that the 2nd derivative of 2x is 0), they can only intersect in one place. i.e. it only has one root for x>0. (and x!=0)

 

[...]

ok, here are some really werid q's

let the function f: (-1,1 ) ---> R be defined by

f(x) = x /( x² - 1) for every x E ( -1,1 )

i) what is the codomain of f?
ii) is f one-to-one? Justify
iii) is f onto? Justify
iv) is f invertible? Justify

the italic is what i dun get..like what it means...

lol..any help is appreciated..thanks

 

[wogboy]

i)
R, as you can see f is defined f: (-1,1) -> R

ii)
solve f'(x) = 0 to find stationary pts of f
f'(x) = (x^2 - 1 - 2x^2)/(x^2 - 1)^2
= -(x^2 + 1)/(x^2 - 1)^2
solving for f'(x)=0 there are no real solutions in the interval [-1,1] and f'(x) < 0 for all x in this interval so f is monotonically increasing on this interval. Also f is continuous on this interval.
So f is one to one on this interval.

iii)
Find the range of f.
Since f is continuous, and f(x) -> infinity as x -> 1, and f(x) -> -infinity as x->-1, for any real k, there must exist a x in the interval [-1,1] such that f(x)=k.
So f is onto.

iv)
Since f is one to one and onto, f is bijective, therefore f is invertible.

 

[...]

haha, thanks wogboy..

but what does those words mean and what they want out of it...

 

[KeypadSDM]

Originally posted by ...
but even then, does that explain sqrt(x) as x-->oo ???

Here's a laymans interpretation:

Now infinity's a long way out there, right?

So:

Lim[x-->oo] (1/Sqrt[x])
= Lim[(Sqrt[x])²-->oo] (1/Sqrt[x])

Number n is approaching a finite number, hence n² is also approaching some finite number. Thus if n² is approaching an infinite number, n must be approaching a non-finite number, hence infinite number.

Let n be Sqrt[x]

Thus:

Lim[(Sqrt[x])²-->oo] (1/Sqrt[x])
=Lim[Sqrt[x]-->oo] (1/Sqrt[x])

Sqrt[x] = y

:. Lim[x-->oo] (1/Sqrt[x])
= Lim[y-->oo] (1/y)
= 0

Another method, much more simple:

Lim[x-->oo] (1/Sqrt[x])
= Sqrt[ Lim[x-->oo] (1/x) ]
= Sqrt[0]
= 0

 

[maniacguy]

I should point out that commuting limits with functions is NOT generally a good idea!! In a "layman's guide" in particular...

 

[...]

hmm..
Evaluate


lim_x-->oo[ x² + 3.cos(5x)] / [x² - 2.sin(4x) ]

somehow this q seem really easy yet i can't make a start on it...

 

[...]

okok, heres one more..but this one might need some explaining

i have 2limts

limit_(x-->oo) sqrt(x² + 4x + 3) - x

and

limit_(x-->-oo) sqrt(x² + 4x + 3) - x


for the 2limits, the value are totally different..hmm..can anyone tell me hwo u would approach the negative infinity part...

 

[wogboy]

for the 2limits, the value are totally different..hmm..can anyone tell me hwo u would approach the negative infinity part...

Here's an EXTREMELY useful tip:

if you see something like:
lim{x -> ?} a(x) - b(x), where a AND b BOTH go off to +infinity (or BOTH go to -infinity, but not mixed) and that a =/= b, use the fact that:

a - b = (a^2 - b^2)/(a+b)

for the limit as x -> -infinity part, you can see that if you put it in the form lim{x->-infinity} a - b, then a goes to +infinity while b goes to -infinty (so you're basically adding two infinities together). The result is of course that the whole limit goes off to +infinity, so the limit does not exist.

 

[wogboy]

but what does those words mean and what they want out of it...

Codomain: This is what is written in the function definition, after the arrow (e.g. f : R -> [0,1] has codomain [0,1] )

One to One (also called injective): This means that every value input into the function f, a unique value is produced. In maths form:

f(a) = f(b) if & only if a = b

Onto (also called surjective): This means that the range of the function f equals the codomain. The range, as you know from HSC maths, is just all the y values covered by a function over it's domain. In maths form:

for every y in the codomain of f, there exists an x in the domain of f such that f(x) = y

Bijective: This means f is both injective and surjective.

Invertible: This means the function f has an inverse function g, such that g(f(x)) = x for all x in the domain of f (global inverse), or some interval of x (local inverse).

Theorem: A function is globally invertible if and only if bijective.

 

[maniacguy]

Alternately (some people find this easier):

One-to-one = if you draw the graph, then when you draw a horizontal line (y=k) it intersects the graph of the function at most once.

Onto = if you draw the graph then when you draw a horizontal line it intersects the graph of the function at least once.

Bijective = if you draw the graph then when you draw a horizontal line it intersects the graph of the function exactly once.

 

[...]

Originally posted by maniacguy
One-to-one = if you draw the graph, then when you draw a horizontal line (y=k) it intersects the graph of the function at most once.

Bijective = if you draw the graph then when you draw a horizontal line it intersects the graph of the function exactly once.

umm..isn't that the same..

at most once, and exactly once..

like u can't have the line crossing the function half a time right

 

[wogboy]

like u can't have the line crossing the function half a time right

But you can have it crossing zero times (i.e. never) e.g. (f: R -> R f(x) = x^2 and the line y= -1)

 

[KeypadSDM]

Or many times...

 

[...]

hmm..

no one can get this?

Originally posted by ...
hmm..
Evaluate


lim_x-->oo[ x² + 3.cos(5x)] / [x² - 2.sin(4x) ]

somehow this q seem really easy yet i can't make a start on it...

 

[CM_Tutor]

Use L'Hopital's Rule

 

[turtle_2468]

alternatively, just divide both sides by x^2 a la 4 Unit
then use well known things like lim of sinx/x is 1, hence lim of sinx/x^2=0

 

[...]

Originally posted by CM_Tutor
Use L'Hopital's Rule

i tried that, but somehow the sin x and cos x is shitting me

seems evil

Originally posted by turtle_2468
alternatively, just divide both sides by x^2 a la 4 Unit
then use well known things like lim of sinx/x is 1, hence lim of sinx/x^2=0

ooh..let me try that

 

[CM_Tutor]

lim_x-->+inf [x² + 3.cos(5x)] / [x² - 2.sin(4x)]
= lim_x-->+inf d/dx[x² + 3.cos(5x)] / d/dx[x² - 2.sin(4x)], using L'Hopital's Rule
= lim_x-->+inf [2x - 15.sin(5x)] / [2x - 8.cos(4x)]
= lim_x-->+inf d/dx[2x - 15.sin(5x)] / d/dx[2x - 8.cos(4x)], using L'Hopital's Rule
= lim_x-->+inf [2 - 75.cos(5x)] / [2 + 32.sin(4x)]

On second thoughts, use turtle_2468's approach...

lim_x-->+inf [x² + 3.cos(5x)] / [x² - 2.sin(4x)] = lim_x-->+inf {[x² + 3.cos(5x)] / x² divided by [x² - 2.sin(4x)] / x²}
= lim_x-->+inf [1 + 3.cos(5x) / x²] / [1 - 2.sin(4x) / x²]
= (1 + 0) / (1 - 0)
= 1