How to Solve this Weird Limit

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  • #1
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How do you solve this kind of limit?

[tex]\lim_{x \to ∞} \left( \frac{x^2 - 1}{x^2 + 2x + 5} \right)^{-2x}[/tex]

Please give me a clue.
 

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  • #2
PeroK
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What about using the logarithm?
 
  • #3
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Thank's for the clue but I'm still thinking it seem too difficult for me. Can you please show me which page from this calculus textbook about solving limit with logarithm?

https://www.sendspace.com/file/khld5g
 
  • #4
PeroK
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Thank's for the clue but I'm still thinking it seem too difficult for me. Can you please show me which page from this calculus textbook about solving limit with logarithm?

https://www.sendspace.com/file/khld5g
I'm not sure that homework helping extends to reading your textbooks!

The natural log is a continuous, one-to-one function. You can use that property to see that (informally):
$$\lim \ln(f(x)) = \ln(\lim f(x))$$
 
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  • #5
Gaussian97
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Alternatively, you can try to put your limit in the form
$$\lim_{y\rightarrow \infty}\left(1+\frac{1}{y}\right)^{y\cdot f(x)}$$
and then use the limit definition of the number ##e##.
You can start by trying to write your quotient ##\frac{x^2 - 1}{x^2 + 2x + 5}## as ##1+\frac{1}{f(x)}##
 
  • #6
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Are all of these within the calculus textbook?
 
  • #8
ehild
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How do you solve this kind of limit?

[tex]\lim_{x \to ∞} \left( \frac{x^2 - 1}{x^2 + 2x + 5} \right)^{-2x}[/tex]

Please give me a clue.
Follow @Gaussian97's hint:
Try to bring it to a function of ##(1+1/y)^y## What is this limit when y goes to infinity?
First divide both the numerator and denominator by x^2. Then neglect the second-order terms.
 
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  • #9
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Please show me the first step, I still don't understand, my brain is blank, too difficult.
 
  • #10
Gaussian97
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Do you know about polynomial division? Any polynomial ##A## (so, of course also the polynomial ##x^2-1##) can be divided by any polynomial ##B## (so, in your case ##x^2+2x+5##), what this means is that you can find polynomials ##Q## and ##R## (with the degree of ##R## lower than the degree of ##B##) such that;
$$A = BQ+R$$
This can be expressed also as
$$\frac{A}{B} = Q+\frac{R}{B}$$
Here you have methods to find ##Q## and ##R## in case you don't know how to do it.
So, the first step is to find those polynomials ##Q## and ##R##, we have advanced you that the polynomial ##Q## is ##Q=1## (which is a polynomial of degree 0). Is now your turn verify this is true and find the polynomial ##R##.

You can start by trying to write your quotient ##\frac{x^2 - 1}{x^2 + 2x + 5}## as ##1+\frac{1}{f(x)}##
Then, the function ##f(x)## that I tell you in this post will be simply the function defined by
$$f(x)=\frac{B}{R}$$ with ##B=x^2+2x+5## and ##R## the polynomian you need to find.
 
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  • #11
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Wow! it's too difficult form me. Please show me the first step.
 
  • #12
Gaussian97
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Ok, I will do an example step by step, try to repeat it with your polynomials:
I will divide ##3x^3+6x^2-3x+6## by ##3x^2-1##

So, the first step, you ignore all the terms except the first one of each polynomial and divide them;
$$\frac{3x^3+6x^2-3x+6}{3x^2-1}\rightarrow \frac{3x^3}{3x^2}=x$$
Now you multiply the divisor by ##x## and subtract from the dividend;
$$(3x^3+6x^2-3x+6) - x\cdot(3x^2-1) = (3x^3+6x^2-3x+6) - (3x^3-x)$$ $$= (3-3)x^3+6x^2-(3-1)x+6 = 6x^2-2x+6$$
Notice that this new polynomial has a lower degree than the original one, and also by construction they fulfil
$$(6x^2-2x+6) + x (3x^2-1) = 3x^3+6x^2-3x+6$$

You can redo these two steps once more, starting with ##6x^2-2x+6##
$$\frac{6x^2-2x+6}{3x^2-1}\rightarrow \frac{6x^2}{3x^2}=2$$
$$6x^2-2x+6 - 2\cdot(3x^2-1) =6x^2-2x+6 - (6x^2-2)$$ $$= (6-6)x^2-2x+(6+2) = -2x+4$$

Now notice that this polynomial has a lower degree that you divisor, so we are done, the polynomials that we want are
$$Q=x+2, \qquad R=-2x+4$$
 
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  • #13
PeroK
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Wow! it's too difficult form me. Please show me the first step.
Re the logarithmic method, the first step is:

$$y(x) = \left( \frac{x^2 - 1}{x^2 + 2x + 5} \right)^{-2x}$$
$$\ln y = 2\frac{\ln(x^2 + 2x + 5) - \ln(x^2 - 1)}{\frac 1 x}$$
 
  • #14
ehild
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Are all of these within the calculus textbook?
Your calculus textbook certainly contains the limit
$$\lim _{x\rightarrow\infty} \left(1+1/x\right)^x=e$$
If you follow my hint in Post #8 you find something similar, in a few simple steps
First divide both the numerator and denominator by x^2. Then neglect the second-order terms with 1/x^2, small with respect to 1 and 2/x if x is big enough. What do you get?
 
  • #15
PeroK
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Wow! it's too difficult form me. Please show me the first step.

Are you doing a rigorous Real Analysis course? If so, the method using the exponential may be quicker but I'm not sure it can easily be made rigorous.

If you have to evaluate this limit rigorously, then I think the logarithmic approach is the way to go.
 
  • #16
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Forgive me but I still don't understand.

Re the logarithmic method, the first step is:

$$y(x) = \left( \frac{x^2 - 1}{x^2 + 2x + 5} \right)^{-2x}$$
$$\ln y = 2\frac{\ln(x^2 + 2x + 5) - \ln(x^2 - 1)}{\frac 1 x}$$

This is like a differentiation using a logarithm. But this is a "limit" problem.

More clue please, I give up.
 
  • #17
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PeroK said:
Re the logarithmic method, the first step is:

$$y(x) = \left( \frac{x^2 - 1}{x^2 + 2x + 5} \right)^{-2x}$$
$$\ln y = 2\frac{\ln(x^2 + 2x + 5) - \ln(x^2 - 1)}{\frac 1 x}$$
This is like a differentiation using a logarithm.
No! All @PeroK did was to take the natural log of both sides.
But this is a "limit" problem.
For the 2nd line of what PeroK wrote, take the limit as ##x \to \infty## on both sides.

Have you looked at your textbook? There should be an example of this technique.
 
  • #18
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Please show me the first step, I still don't understand, my brain is blank, too difficult.
Wow! it's too difficult form me. Please show me the first step.
Forgive me but I still don't understand.
@askor, you've been given lots of hints, but have not shown any effort on your part. If you are not able to make some progress in this problem with the help you've been given, I'm going to close this thread.
 
  • #19
PeroK
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Forgive me but I still don't understand.

This is like a differentiation using a logarithm. But this is a "limit" problem.

More clue please, I give up.

As @Mark44 pointed out, all I did was take the natural log and do some useful manipulation. That wasn't just a clue, that was actually quite a large first step I did for you.

The fact that you don't understand at all what I'm doing suggests that this material is too advanced for you. You don't understand the general method, and you don't understand the algebraic manipulations.

This question is not easy, so I suggest you find some more basic questions first. You should learn the logarithmic method for limits. Also, learn l'Hopital's rule. And try to improve your algebra.
 
  • #21
lurflurf
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Suppose we have
$$\lim_{x\rightarrow\infty} a^b$$
with
$$\lim_{x\rightarrow\infty} a=1$$
$$\lim_{x\rightarrow\infty} b=\infty$$

$$a^b=\exp(b\log(a))$$
$$a^b=\exp\left( b (a-1)) \frac{\log(a)-\log(1)}{a-1}\right)$$

Where to go from here?
 
  • #22
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The fact that you don't understand at all what I'm doing suggests that this material is too advanced for you. You don't understand the general method, and you don't understand the algebraic manipulations.
The same thought crossed my mind ...

Please, no more clues until @askor responds.
 

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