Ok so need help with this question. Sadly it's in the textbook so all I have is the answer. The problem I have is I don't see how the x terms cancel.
x→∞lim[(x2+3x)21−x] Any help would be appreciated. If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.
Ok so need help with this question. Sadly it's in the textbook so all I have is the answer. The problem I have is I don't see how the x terms cancel.
x→∞lim[(x2+3x)21−x] Any help would be appreciated. If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.
Do you need to prove the limit or just find it.
If it is the latter just try some big numbers for x such as 1000000.
Ok so need help with this question. Sadly it's in the textbook so all I have is the answer. The problem I have is I don't see how the x terms cancel.
x→∞lim[(x2+3x)21−x] Any help would be appreciated. If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.
Does your textbook give any similar examples? It's difficult to give a hint for this without explaining exactly how to do it if you haven't seen it before.
Does your textbook give any similar examples? It's difficult to give a hint for this without explaining exactly how to do it if you haven't seen it before.
Nope, all the examples given are of functions divided by another function. That's why this one confused me.
Ok so need help with this question. Sadly it's in the textbook so all I have is the answer. The problem I have is I don't see how the x terms cancel.
x→∞lim[(x2+3x)21−x] Any help would be appreciated. If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.
[br]x2+3x−x=((x+23)2−49)1/2−x[br] For large x, the −49 is insignificant thus the equation becomes: ((x+23)2)21−x=x+23−x=23 Thus the answer is 1.5
Thanks! I didn't know you could expand it in that form though, it makes it more obvious why you can find the limit though, and makes the answer downright obvious.
Ok so need help with this question. Sadly it's in the textbook so all I have is the answer. The problem I have is I don't see how the x terms cancel.
x→∞lim[(x2+3x)21−x] Any help would be appreciated. If anyone wants to know its the AQA MFP3 book provided on the internet. Page 24 Excersise 1D Question 5.
Now you've seen a few suggestions, here is the cleanest way to attack this.
As MrM says, multiply what you've got by:
x2+3x+xx2+3x+x
You then end up with:
x2+3x+x3x or 1+(3/x)+13
You can now apply the algebra of limits directly, as x−>∞, 3/x->0, 1+(3/x)->1, the square root tends to 1 and the denominator tends to 2.
Hence the limit is 3/2.
This method avoids dealing with infinite expansions or using any claims about terms being "negligible" without proof
Now you've seen a few suggestions, here is the cleanest way to attack this.
As MrM says, multiply what you've got by:
x2+3x+xx2+3x+x
You then end up with:
x2+3x+x3x or 1+(3/x)+13
You can now apply the algebra of limits directly, as x−>∞, 3/x->0, 1+(3/x)->1, the square root tends to 1 and the denominator tends to 2.
Hence the limit is 3/2.
This method avoids dealing with infinite expansions or using any claims about terms being "negligible" without proof
I tend to avoid the negligible ones anyway. Not really sure when you're allowed to use it so.... I still used the infinite expansion. It's a bit of extra writing but that's what they're expecting you to do so I though, Why not?
I tend to avoid the negligible ones anyway. Not really sure when you're allowed to use it so.... I still used the infinite expansion. It's a bit of extra writing but that's what they're expecting you to do so I though, Why not?
If that's what they teach you for A level then that's fine - technically you can't assume that you can take an infinite limit inside an infinite series and get the right answer, and the method I provided avoids the need for infinite series altogether, but go with whatever's allowed by your syllabus