Limit Problem with Infinity

[Jason76]

EDITED

\(\displaystyle \dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}\)

Now divide everything by the highest index in the denominator

\(\displaystyle \dfrac{\lim}{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}\)

Now simplify

\(\displaystyle \dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + ? - 4}\) How do I evaluate the term in the middle of the denominator? I can sort of see now what is going on.

This should come out to \(\displaystyle -\dfrac{1}{4}\)

 

[DrPhil]

\(\displaystyle \displaystyle\lim_{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}\)

Now divide everything by the highest index in the denominator

\(\displaystyle \displaystyle \lim_{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}\)

Are you planning to simplify that? Shall I wait for you to finish editing before replying?

EDIT following your edit:
Simplify BEFORE taking the limit.

What is \(\displaystyle x^2/x^2\) ?

REEDIT - please let us know whan you are through editing your post.

 

[Jason76]

Are you planning to simplify that? Shall I wait for you to finish editing before replying?

EDIT following your edit:
Simplify BEFORE taking the limit.

What is \(\displaystyle x^2/x^2\) ?

Now original post edited

 

[Deleted member 4993]

x/x² = 1/x

 

[Jason76]

\(\displaystyle \dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}\)

Now divide everything by the highest index in the denominator

\(\displaystyle \dfrac{\lim}{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}\)

\(\displaystyle \dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4}\) - :!: See the \(\displaystyle x^{-1}\) in the bottom?

\(\displaystyle \dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4}\) - :!: This becomes \(\displaystyle x^{1}\) in the top. But how to evaluate to get \(\displaystyle -\dfrac{1}{4}\)

 

[MarkFL]

...
\(\displaystyle \dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4}\) - :!: This becomes \(\displaystyle x^{1}\) in the top. But how to evaluate to get \(\displaystyle -\dfrac{1}{4}\)

No, that is not a valid algebraic manipulation, you cannot take an addend "to the top" like that, only a factor.

If \(\displaystyle \displaystyle \lim_{x\to\infty}\frac{1}{x^2}=0\) then what can you say about \(\displaystyle \displaystyle \lim_{x\to\infty}\frac{1}{x}\)?

 

[Jason76]

No, that is not a valid algebraic manipulation, you cannot take an addend "to the top" like that, only a factor.

If \(\displaystyle \displaystyle \lim_{x\to\infty}\frac{1}{x^2}=0\) then what can you say about \(\displaystyle \displaystyle \lim_{x\to\infty}\frac{1}{x}\)?

Sorry, I have no idea. I do see that we have to divide \(\displaystyle \dfrac{x}{x^{2}}\)(in the denominator). What do we do in that situation? How does that relate to the final answer?

 

[MarkFL]

How did you determine \(\displaystyle \displaystyle \lim_{x\to\infty}\frac{1}{x^2}=0\)?

Can you apply similar reasoning to \(\displaystyle \displaystyle \lim_{x\to\infty}\frac{1}{x}\)?

 

[lookagain]

\(\displaystyle \displaystyle\lim_{x \to \infty} \ \dfrac{x^{2} - 4}{2 + x - 4x^{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)You should see if the numerator and denominator would factor so that you could cancel (divide out) any factors between them.
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)Here, it turns out that there are no factors to cancel (divide out).



Suggestion: Rewrite the expression so that both polynomials have descending degrees for their terms:

\(\displaystyle \displaystyle\lim_{x \to \infty} \ \dfrac{x^{2} - 4}{- 4x^{2} + x + 2} \ = \)

Divide all terms by the term with the highest degree, which is \(\displaystyle \ x^2: \)


\(\displaystyle \displaystyle\lim_{x \to \infty} \ \dfrac{1 \ - \ \dfrac{4}{x^2}}{-4 \ + \ \dfrac{1}{x} \ + \ \dfrac{2}{x^2}} \ =\)

The limit as x approaches infinity of each constant is itself. The limit as x approaches infinity of each of the three fractions is 0.

\(\displaystyle \ \)Do you know why that's true (regarding the second sentence)?




\(\displaystyle \ \ \ \)Replace each of those three fractions with 0, and remove the limit notation that has been to the left of the larger fractional expression:



\(\displaystyle \dfrac{1 \ - \ 0}{-4 \ + \ 0 \ + 0} \ = \ \ \ \ \ \ \) <----- This step is wrong if you still have the limit notation to the left of this fraction.


\(\displaystyle \boxed{ \ \dfrac{-1}{4} \ \ \ or \ \ \ -\dfrac{1}{4}}\)

 

[Jason76]

Oh, I see now. \(\displaystyle \dfrac{x}{x^{2}} = x^{-1} = \dfrac{1}{x}\) This comes out to zero when infinity is plugged in.

 

[lookagain]

Oh, I see now. \(\displaystyle \dfrac{x}{x^{2}} = x^{-1} = \dfrac{1}{x} \ \ \ \ \ \ \) Yes.


This comes out to zero when infinity is plugged in. \(\displaystyle \ \ \ \ \ \)No, it equals 0 as x approaches infinity.

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