limits help (1 Viewer)

[starryblue]

i dont understand how to do these qu,

1) limit x-> 7 (2x+3)/(x-7)
2) the function f(x)=(4x^2-9)/(2x-3) is not continuous throughout its domain. redefine f(x) such that it is continuous throughout its domain
3) limit x approaches infinity (x^3-2x^2+3x+4)/(3x+5)
4) limit x app infinity (sqrt (2x+3))-sqrt(2x))


and a qu on first principle diff

y=4x sqrt x

thanks (god, this took forever to type on the tablet ==)

 

[Carrotsticks]

Question 1: Are you sure you typed it out correctly?

Question 2: The numerator is the difference of two squares, and the bottom cancels with one of the terms. To be picky, writing it as a 'new function' doesn't make the discontinuity disappear, but I presume your teacher just wants you to just factorise it.

Question 3: The top clearly approaches infinity faster than the bottom, so the limit is just infinity.

Question 4: Rationalise the numerator, and the answer will fall out (the limit is zero). Another way of thinking about it is that for large values of x, 2x+3 ~ 2x, so we are essentially doing 2x - 2x, which is zero.

 

[iBibah]

Question 1: Are you sure you typed it out correctly?

Question 2: The numerator is the difference of two squares, and the bottom cancels with one of the terms. To be picky, writing it as a 'new function' doesn't make the discontinuity disappear, but I presume your teacher just wants you to just factorise it.

Question 3: The top clearly approaches infinity faster than the bottom, so the limit is just infinity.

Question 4: Rationalise the numerator, and the answer will fall out (the limit is zero). Another way of thinking about it is that for large values of x, 2x+3 ~ 2x, so we are essentially doing 2x - 2x, which is zero.

Isn't it plus or minus infinity?

 

[Carrotsticks]

Isn't it plus or minus infinity?

Yep it is as it stands right now, but usually questions have nice finite values.

 

[COLDBOY]

i dont understand how to do these qu,

1) limit x-> 7 (2x+3)/(x-7)
2) the function f(x)=(4x^2-9)/(2x-3) is not continuous throughout its domain. redefine f(x) such that it is continuous throughout its domain
3) limit x approaches infinity (x^3-2x^2+3x+4)/(3x+5)
4) limit x app infinity (sqrt (2x+3))-sqrt(2x))


and a qu on first principle diff

y=4x sqrt x

thanks (god, this took forever to type on the tablet ==)

1) (2x +3)/(x-7) question is incorrect!
2)f(x) = (4x^2 - 9)/(2x-3) = (2x-3)(2x+3)/(2x-3) = 2x+3.--> Domain All real x (hence continuous)
3) (x^3 - 2x^2 + 3x + 4)/(3x+5) --> (1/3)x^2 + (11/25)x + 4/5 --> infinity
4)limit as x --> infinity, (sqrt (2x+3))-sqrt(2x))--> infinity - infinity --> 0 (sqrt(infinity) --> infinity)

y = 4xsqrt(x) = 4*x^(3/2)

now

f'(x) = y' = lim(h-> 0) (4(x+h)^(3/2) - x^(3/2))/h
= 4* lim(h-> 0) ((x+h)^(3/2) - x^(3/2))/h
= 4* lim(h-> 0) ((x^(3/2) + (3/2)x^(3/2-1)*h +...(terms that keep going including h with powers greater than or equal to 2) - x^(3/2))/h
= 4* lim(h-> 0)((3/2)x^(1/2) + ( bunch of terms with powers of h greater than equal to 2)/h i.e. limit((h^n)/h) --> 0 were n>=2 h--> 0)
= 4*(3/2)x^(1/2)
=6sqrt(x)

 

[iBibah]

1) (2x +3)/(x-7) question is incorrect!
2)f(x) = (4x^2 - 9)/(2x-3) = (2x-3)(2x+3)/(2x-3) = 2x+3.--> Domain All real x (hence continuous)
3) (x^3 - 2x^2 + 3x + 4)/(3x+5) --> (1/3)x^2 + (11/25)x + 4/5 --> infinity
4)limit as x --> infinity, (sqrt (2x+3))-sqrt(2x))--> infinity - infinity --> 0 (sqrt(infinity) --> infinity)

y = 4xsqrt(x) = 4*x^(3/2)

now

f'(x) = y' = lim(h-> 0) (4(x+h)^(3/2) - x^(3/2))/h
= 4* lim(h-> 0) ((x+h)^(3/2) - x^(3/2))/h
= 4* lim(h-> 0) ((x^(3/2) + (3/2)x^(3/2-1)*h +...(terms that keep going including h with powers greater than or equal to 2) - x^(3/2))/h
= 4* lim(h-> 0)((3/2)x^(1/2) + ( bunch of terms with powers of h greater than equal to 2)/h i.e. limit((h^n)/h) --> 0 were n>=2 h--> 0)
= 4*(3/2)x^(1/2)
=6sqrt(x)

Isn't it plus or minus infinity?

Yep it is as it stands right now, but usually questions have nice finite values.

.

 

[planino]

2)f(x) = (4x^2 - 9)/(2x-3) = (2x-3)(2x+3)/(2x-3) = 2x+3.--> Domain All real x (hence continuous)

Wouldn't the domain be?

All real x
x not equal to 3/2

since 2x - 3 can't equal 0 (even though it was cancelled out) as it's in the denominator? So I think the function would look like y=2x+3 with an open circle at x=3/2

 

[Carrotsticks]

Wouldn't the domain be?

All real x
x not equal to 3/2

since 2x - 3 can't equal 0 (even though it was cancelled out) as it's in the denominator? So I think the function would look like y=2x+3 with an open circle at x=3/2

That is correct, which was why I was a bit apprehensive about the question in the first place, because it seems to assume that just because we 'cancel' out the denominator, the discontinuity vanishes, which is NOT true.

 

[iBibah]

It reminds me y=x^3/x.
It's the well known y=x^2 graph however an open circle at the origin, but some people seem to think you can simplify it before drawing (you can if it makes it easier but never forget the original function, because that's where you get your domain from).

 

[Leffife]

1) REWRITE THE QUESTION, YOU MUST HAVE DONE SOMETHING WRONG.
2) Factorise it: f(x) = (2x+3)(2x-3)/(2x-3). Do the cancelling. f(x) = (2x+3) from here you should realise the domain is continuous - i.e. linear equation, thus D: All Real Values of x

And the rest should be done easily.

 

[Shadowdude]

1) REWRITE THE QUESTION, YOU MUST HAVE DONE SOMETHING WRONG.
2) Factorise it: f(x) = (2x+3)(2x-3)/(2x-3). Do the cancelling. f(x) = (2x+3) from here you should realise the domain is continuous - i.e. linear equation, thus D: All Real Values of x

And the rest should be done easily.

Why? There isn't a limit because the function tends to positive or negative infinity. You can't assume the question is wrong, plenty of functions tend to infinity, like the limit of x as x to infinity, or x^2, or x^3...

 

[twinklegal19]

That is correct, which was why I was a bit apprehensive about the question in the first place, because it seems to assume that just because we 'cancel' out the denominator, the discontinuity vanishes, which is NOT true.

can "redefining a function" to make it continuous throughout its domain possibly involve forming a piecemeal function instead? Such as



just a thought



P.S. also, took ages to latex that perfectly lol

 

[SpiralFlex]

can "redefining a function" to make it continuous throughout its domain possibly involving forming a piecemeal function instead? Such as



just a thought



P.S. also, took ages to latex that perfectly lol

LOL! Hahahahahaha

 

[twinklegal19]

LOL! Hahahahahaha

it's the same thing as piecewise...

w/e

 

[RealiseNothing]

Hmmm.

Define the set of functions that diverge to as

Similarly define the set of functions that converge to some value such that there exists some value where as:



Question, does one of the above sets contain more elements than the other? If so, which one.

 

[Fus Ro Dah]

Hmmm.

Define the set of functions that diverge to as

Similarly define the set of functions that converge to some value such that there exists some value where as:



Question, does one of the above sets contain more elements than the other? If so, which one.

One could consider divergence to be convergence to infinity...

 

[RealiseNothing]

One could consider divergence to be convergence to infinity...

Hence the "set of functions that converge to some value such that there exists some value where ".

Unless you are referring to something else?

 

[Fus Ro Dah]

Hence the "set of functions that converge to some value such that there exists some value where ".

Unless you are referring to something else?

Didn't read that part, sorry.