Finding limits without use of l'Hôpital's rule.

  • Thread starter Mutaja
  • Start date
  • #1
Mutaja
239
0

Homework Statement



Hi.

I have a problem finding the limit of two different problems - without use of l'Hôpital's rule. I only know how to do this with use of the l'Hôpital's rule, therefore I'm seeking help to solve this problem.


Homework Equations



The problems are:

Determine the limits without using l'Hopital's rule:

lim (x-2)/((√(x^2-3)-x/2)
x->2

lim (2x-√(4x^2+3x))
x->∞

The Attempt at a Solution



First thing first, the 1st problem.

I do not have any attempt at a solution yet, as I do not know how to go about to solve this problem. I should add that I have solved this problem graphically, and I know that the answer should be 2/3.

Should I attempt to simplify the expression (I'm a bit rusty on that, unfortunately), factorize the expression or how should I go about to solve this?

I'm a bit clueless here, but I have my books open, ready to make an effort into completing this problem.

Thanks for any input.
 

Answers and Replies

  • #2
Mutaja
239
0
I do know that I'll have to do something with the expression in order to be able to cancel something out with each other.

My best attempt so far is to multiply both the numerator and denominator by √(x^2-3), but I'm not too sure on how I write that out.
 
  • #3
Mutaja
239
0
Can anyone at least point me towards how I can find the solution to this problem myself?
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
20,731
11,565
I don't know, how to do the work without de L'Hospital's rule. The only idea I have is to use series expansions, but in fact that's also de L'Hosptital's rule, but it's hidden behind its proof ;-).

E.g., for the first problem
[tex]\sqrt{x^2-3}-\frac{x}{2}=\frac{3}{2}(x-2)+\mathcal{O}[(x-2)^2].[/tex]
 
  • #5
Mutaja
239
0
I don't know, how to do the work without de L'Hospital's rule. The only idea I have is to use series expansions, but in fact that's also de L'Hosptital's rule, but it's hidden behind its proof ;-).

E.g., for the first problem
[tex]\sqrt{x^2-3}-\frac{x}{2}=\frac{3}{2}(x-2)+\mathcal{O}[(x-2)^2].[/tex]

Thanks a lot for showing interest in my topic, even if I haven't got any solid work to analyze.

I do, however, doubt that this is how I should solve this problem, because we haven't learned anything similar to this I'm afraid. To be honest, I don't know what technique I'm supposed to use for this problem.

Would it be possible to factorize/simplify the expression, cancel out identical expressions, and then swap 2 for x, and get 2/3 as an answer? I've tried, but I can't figure out how to factorize/simplify it. I'm currently looking at how to simplify and factorize expressions like this, so I post any solutions or progress I find.
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
20,731
11,565
Ok, for the first problem, you can also try to rationalize the denominator, using
[tex](\sqrt{x^2-3}-x/2)(\sqrt{x^2-3}+x/2)=(x^2-3)-x^2/4=3/4 x^2-3.[/tex]
 
  • #7
Skins
74
4
For the first question change the denominator to

[tex]\sqrt{x^2 - 3} - \frac{x}{2} [/tex]

Then multiply the top and bottom by

[tex]\sqrt{x^2 - 3} + \frac{x}{2} [/tex]

Then by doing some factorization of the top and bottom and some cancellations of terms you should get the desired result. I scribbled it out an a sheet of paper here and it seems to works out.

I haven't had a chance to try the second one yet... Maybe when I get back I'll give it a shot. In any event for the first no series expansions or approximations, etc. are required. Just some good ol; fashioned elementary algebra.

Hope this helps.
 
  • #8
36,194
8,179

Homework Statement



Hi.

I have a problem finding the limit of two different problems - without use of l'Hôpital's rule. I only know how to do this with use of the l'Hôpital's rule, therefore I'm seeking help to solve this problem.


Homework Equations



The problems are:

Determine the limits without using l'Hopital's rule:

lim (x-2)/((√(x^2-3)-x/2)
x->2

lim (2x-√(4x^2+3x))
x->∞

The Attempt at a Solution



First thing first, the 1st problem.

I do not have any attempt at a solution yet, as I do not know how to go about to solve this problem. I should add that I have solved this problem graphically, and I know that the answer should be 2/3.

Should I attempt to simplify the expression (I'm a bit rusty on that, unfortunately), factorize the expression or how should I go about to solve this?

I'm a bit clueless here, but I have my books open, ready to make an effort into completing this problem.

Thanks for any input.

For the second problem, multiply numerator and denominator by 1 in the form of 2x + √(4 x2 + 3x) over itself. After doing some careful factoring of the denominator, you should be able to evaluate your limit.

BTW, it's against Physics Forums rules to "bump" your thread sooner than 24 hours after you post it.
 
  • #9
Skins
74
4
Hint... for the second limit rationalize by multiplying by

[tex] \frac{2x + \sqrt{4x^2 + 3x}}{2x + \sqrt{4x^2 + 3x}}[/tex]

So...

[tex](2x - \sqrt{4x^2+3x})\frac{2x + \sqrt{4x^2 + 3x}}{2x + \sqrt{4x^2 + 3x}} [/tex]

Then proceed from there with. Factor when possible and simplify radical expressions. Give it a try and let us know how it works out.
 
Last edited:
  • #10
Mutaja
239
0
For the first question change the denominator to

[tex]\sqrt{x^2 - 3} - \frac{x}{2} [/tex]

Then multiply the top and bottom by

[tex]\sqrt{x^2 - 3} + \frac{x}{2} [/tex]

Then by doing some factorization of the top and bottom and some cancellations of terms you should get the desired result. I scribbled it out an a sheet of paper here and it seems to works out.

I haven't had a chance to try the second one yet... Maybe when I get back I'll give it a shot. In any event for the first no series expansions or approximations, etc. are required. Just some good ol; fashioned elementary algebra.

Hope this helps.

Thanks a lot for this! I've been away for awhile, but I'm back now for a couple of hours of math problem solving.

This is what I've gotten so far.

[itex]\frac{(x-2)(\sqrt{x^2-3}+\frac{x}{2})}{(\sqrt{x^2-3}-\frac{x}{2})(\sqrt{x^2-3}+\frac{x}{2})}[/itex]

= [itex]\frac{(x-2)(\sqrt{x^2-3}+\frac{x}{2})}{(\sqrt{(x^2-3) ^2}-\frac{x}{2}^2}[/itex] <--- the (x/2)^2 should be like this. Not (x^2/2).

= [itex]\frac{(x-2)(\sqrt{x^2-3)}+\frac{x}{2})}{(x^2-3)-(\frac{x^2}{4})}[/itex]

Where am I going wrong?

Thanks a lot for all the help, and I apologize for not following the rules and spamming the thread earlier. I will look at the other inputs when I'm done with this first problem. I can't handle several problems at once I'm afraid.
 
  • #11
36,194
8,179
Your work looks fine so far. To continue, combine the two x2 terms in the denominator and factor the resulting binomial.
 
  • #12
Mutaja
239
0
Your work looks fine so far. To continue, combine the two x2 terms in the denominator and factor the resulting binomial.

I thought I was heading for trouble, thank you!

[itex]\frac{(x-2)(\sqrt{x^2-3}+\frac{x}{2})}{\frac{3x^2}{4}-3}[/itex]

[itex]\frac{4(x-2)(\sqrt{x^2-3}+\frac{x}{2})}{3(x+2)(x-2)}[/itex] (3x^2-12 rewritten)

[itex]\frac{4(\sqrt{x^2-3}+\frac{x}{2})}{3(x+2)}[/itex]

[itex]\frac{2\sqrt{x^2-3}+x}{3(x+2)}[/itex]

When I now insert 2 for x, I get [itex]\frac{2}{3}[/itex], thank you for all your help :) If this looks ok now, I will be moving on to the next problem.

By the way, it's a paper I'm doing atm. Can I post my next problem in the same post, or do I have to make a new post for another problem?
 
  • #13
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
Make a new thread.
 
  • #14
Mutaja
239
0
For the second problem, multiply numerator and denominator by 1 in the form of 2x + √(4 x2 + 3x) over itself. After doing some careful factoring of the denominator, you should be able to evaluate your limit.

This is what I've got so far.


(2x-[itex]\sqrt{4x^2+3x})[/itex][itex]\frac{2x+\sqrt{4x^2+3x}}{2x+\sqrt{4x^2+3x}}[/itex]

= [itex]\frac{4x^2-4x^2+3x}{2x+\sqrt{4x^2+3x}}[/itex]

=[itex]\frac{3x}{2x+\sqrt{4x^2+3x}}[/itex]

Have I gone by it the wrong way? I feel like in order to continue this problem, I need to do the same thing in the denominator now - multiply it with itself or itself with - instead or +.

Thanks for any input.
 
  • #15
Dick
Science Advisor
Homework Helper
26,263
619
This is what I've got so far.


(2x-[itex]\sqrt{4x^2+3x})[/itex][itex]\frac{2x+\sqrt{4x^2+3x}}{2x+\sqrt{4x^2+3x}}[/itex]

= [itex]\frac{4x^2-4x^2+3x}{2x+\sqrt{4x^2+3x}}[/itex]

=[itex]\frac{3x}{2x+\sqrt{4x^2+3x}}[/itex]

Have I gone by it the wrong way? I feel like in order to continue this problem, I need to do the same thing in the denominator now - multiply it with itself or itself with - instead or +.

Thanks for any input.

##\sqrt{4x^2+3x}=\sqrt{x^2 (4+3/x)}=\sqrt{x^2} \sqrt{4+3/x}##. See where to go from here?
 
  • #16
Mutaja
239
0
##\sqrt{4x^2+3x}=\sqrt{x^2 (4+3/x)}=\sqrt{x^2} \sqrt{4+3/x}##. See where to go from here?

Unfortunately, no. I understand that you're rewriting the root in the denominator, but when I try to continue what you've done, I just end up with wanting to multiply by x, and that's just a step backwards, isn't it?

Thanks a lot for your reply, I really appreciate the help.
 
  • #17
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
How much does 3/x measure relative to 4 when x is really big?
 
  • #18
Mutaja
239
0
How much does 3/x measure relative to 4 when x is really big?

0, I think, but I can't see how that's helping me right now, sorry!

Hmm, can I do it like this?

[itex]\frac{3x}{\sqrt{x^2}\sqrt{4+\frac{3}{x}}}[/itex]

[itex]\frac{3x}{x\sqrt{4+\frac{3}{x}}}[/itex]

Well, and then get something, something, [itex]\frac{3}{4}[/itex]

I'm not sure what I was thinking...
 
  • #19
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
Well, 3/x becomes totally irrelevant towards 4, so that the MAIN contribution from your second term will be [itex]\sqrt{x^{2}}\sqrt{4}=2x[/itex]
Since this main contribution does NOT cancel with the first term in the denominator (ALSO 2x), the total main component of the denominator=2x+2x=4x (in addition, you have some tiny nonsense that goes to zero as x goes to infinity)
 
  • #20
Mutaja
239
0
Well, 3/x becomes totally irrelevant towards 4, so that the MAIN contribution from your second term will be [itex]\sqrt{x^{2}}\sqrt{4}=2x[/itex]
Since this main contribution does NOT cancel with the first term in the denominator (ALSO 2x), the total main component of the denominator=2x+2x=4x (in addition, you have some tiny nonsense that goes to zero as x goes to infinity)

By following this logic, I get that the limit as x approaches infinity is 3/2. My graph tells me it should be -3/4.
 
  • #21
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
First off:
In post 14, you lost a minus sign in tyhe numerator, it should read -3x, rather than 3x.
Secondly, also from post 14, you have TWO terms in the denominator, the first 2x, the other the root expression to be approximated by 2x.

Thus, you get: -3x/(2x+2x)=-3/4
 
  • #22
Mutaja
239
0
First off:
In post 14, you lost a minus sign in tyhe numerator, it should read -3x, rather than 3x.
Secondly, also from post 14, you have TWO terms in the denominator, the first 2x, the other the root expression to be approximated by 2x.

Thus, you get: -3x/(2x+2x)=-3/4

Ah, didn't notice that.

All done, and thanks for helping me out! I've learned a lot about this.
 

Suggested for: Finding limits without use of l'Hôpital's rule.

  • Last Post
Replies
3
Views
1K
Replies
2
Views
6K
  • Last Post
Replies
4
Views
3K
Replies
8
Views
1K
Replies
2
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
7
Views
835
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
4
Views
15K
Replies
3
Views
1K
Top