Finding and Proving the limit as x approaches infinity.
I need to evaluate the limit as x approaches infinity of $dfrac{1-x^2}{x-2}$ and then prove it.
I have found the limit to be negative infinity. Now to prove it I believe I need to find two subsequences that diverge to two different results and therefore prove that the limit actually does not exist. Am I thinking correctly? If so I am stuck on what subsequences to choose.
calculus
edited Nov 30 at 16:24
amWhy
191k28224439
asked Nov 30 at 16:21
Scott Dye
11
add a comment |
I need to evaluate the limit as x approaches infinity of $dfrac{1-x^2}{x-2}$ and then prove it.
I have found the limit to be negative infinity. Now to prove it I believe I need to find two subsequences that diverge to two different results and therefore prove that the limit actually does not exist. Am I thinking correctly? If so I am stuck on what subsequences to choose.
calculus
edited Nov 30 at 16:24
amWhy
191k28224439
asked Nov 30 at 16:21
Scott Dye
11
4
What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
– T. Bongers
Nov 30 at 16:27
I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
– Scott Dye
Nov 30 at 19:00
add a comment |
I need to evaluate the limit as x approaches infinity of $dfrac{1-x^2}{x-2}$ and then prove it.
I have found the limit to be negative infinity. Now to prove it I believe I need to find two subsequences that diverge to two different results and therefore prove that the limit actually does not exist. Am I thinking correctly? If so I am stuck on what subsequences to choose.
calculus
edited Nov 30 at 16:24
amWhy
191k28224439
asked Nov 30 at 16:21
Scott Dye
11
I need to evaluate the limit as x approaches infinity of $dfrac{1-x^2}{x-2}$ and then prove it.
I have found the limit to be negative infinity. Now to prove it I believe I need to find two subsequences that diverge to two different results and therefore prove that the limit actually does not exist. Am I thinking correctly? If so I am stuck on what subsequences to choose.
calculus
calculus
edited Nov 30 at 16:24
amWhy
191k28224439
asked Nov 30 at 16:21
Scott Dye
11
edited Nov 30 at 16:24
amWhy
191k28224439
asked Nov 30 at 16:21
Scott Dye
11
edited Nov 30 at 16:24
amWhy
191k28224439
edited Nov 30 at 16:24
amWhy
191k28224439
edited Nov 30 at 16:24
amWhy
191k28224439
191k28224439
asked Nov 30 at 16:21
Scott Dye
11
asked Nov 30 at 16:21
Scott Dye
11
asked Nov 30 at 16:21
Scott Dye
11
11
4
What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
– T. Bongers
Nov 30 at 16:27
I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
– Scott Dye
Nov 30 at 19:00
add a comment |
4
What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
– T. Bongers
Nov 30 at 16:27
I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
– Scott Dye
Nov 30 at 19:00
4
4
What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
– T. Bongers
Nov 30 at 16:27
What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
– T. Bongers
Nov 30 at 16:27
I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
– Scott Dye
Nov 30 at 19:00
I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
– Scott Dye
Nov 30 at 19:00
add a comment |
1 Answer
1
active
oldest
votes
Hint:
For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
$$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$
I would rather try the following approach (first divide by $x$)
$$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$
answered Nov 30 at 16:57
Dr. Mathva
921316
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
Hint:
For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
$$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$
I would rather try the following approach (first divide by $x$)
$$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$
answered Nov 30 at 16:57
Dr. Mathva
921316
add a comment |
Hint:
For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
$$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$
I would rather try the following approach (first divide by $x$)
$$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$
answered Nov 30 at 16:57
Dr. Mathva
921316
add a comment |
Hint:
For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
$$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$
I would rather try the following approach (first divide by $x$)
$$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$
answered Nov 30 at 16:57
Dr. Mathva
921316
Hint:
For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
$$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$
I would rather try the following approach (first divide by $x$)
$$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$
answered Nov 30 at 16:57
Dr. Mathva
921316
answered Nov 30 at 16:57
Dr. Mathva
921316
answered Nov 30 at 16:57
Dr. Mathva
921316
answered Nov 30 at 16:57
Dr. Mathva
921316
921316
add a comment |
add a comment |
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4
What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
– T. Bongers
Nov 30 at 16:27
I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
– Scott Dye
Nov 30 at 19:00