limits (1 Viewer)

[crammy90]

when exactly do we use this "special limits"
lim/(x-->infinity) 1/x = 0
e.g.
(x^2 + 2x - 3)/(3x^2 - x + 2)
i just learnt it and basically ive realised that you just get the largest number with x in it in the denominator (i.e. 3x^2) and divide everything by this. This leaves one number in the denominator without an x and on the bottom you get a 1 (as u divided 3x^2 by 3x^2) and so the answer is just the number on top?
is this pointless to learn lol could some1 give me a past hsc quedstion where we have had to know it
thanks

 

[Js^-1]

Its just logic. If you have a fraction where you change the number on the bottom, making it bigger and bigger, the fraction gets smaller and smaller.

As for the question, all you have to do is divide every term in the fraction by the highest power of x. Then, anything that is a fraction with x on the bottom tends to zero as x tends to infinity.

Eg. lim_x-->Infinity (x³ + 2x² - 4x⁴) / (2x⁴ + 7x)

Divide every term by x⁴​

= lim_x-->Infinity ( 1/x + 2/x² - 4) / ( 2 + 7/x³)

Everything of the form a/xⁿ tends to zero on account of the limit. ​

= ( 0 + 0 - 4) / ( 2 + 0 )
= - 4 / 2
= - 2

Basically, it's the concept you need to know.
As for whether or not they will ask you a question like this in the exam, I have no idea.

 

[crammy90]

Its just logic. If you have a fraction where you change the number on the bottom, making it bigger and bigger, the fraction gets smaller and smaller.

As for the question, all you have to do is divide every term in the fraction by the highest power of x. Then, anything that is a fraction with x on the bottom tends to zero as x tends to infinity.

Eg. lim_x-->Infinity (x³ + 2x² - 4x⁴) / (2x⁴ + 7x)

Divide every term by x⁴​

= lim_x-->Infinity ( 1/x + 2/x² - 4) / ( 2 + 7/x³)

Everything of the form a/xⁿ tends to zero on account of the limit. ​

= ( 0 + 0 - 4) / ( 2 + 0 )
= - 4 / 2
= - 2

Basically, it's the concept you need to know.
As for whether or not they will ask you a question like this in the exam, I have no idea.

thanks heaps. So you wouldnt need to do this to get the limit of a geometric progression of series? it would just be like a 2marker asking you to find its limit or somethin?

 

[Js^-1]

Yeh, you might get a question in mechanics, something like:

The displacement x of an object at time t is given by

x = 7 ( 1 - e^-2t )

What is the limiting value of this displacement as t --> Infinity?

In this case, you just say:

x_max = lim_{t --> Infinity} 7 - 7/e^2t
x_max = 7 - 0
x_max = 7 Units
Since e^x --> Infinity as x--> Infinity
.: 7/e^x --> 0 as x--> Infinity.