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I am looking to clarify the approach to solve the following limits:

f(x) = $\frac {4x - 2x^2}{x^2 - 4}$ find $\lim \limits_{x\to \infty}$ f(x)

and

$\lim \limits_{x\to 2}$ f(x)

To solve for $\lim \limits_{x\to \infty}$ f(x) I can divide by the largest power and arrive at:

$\lim \limits_{x\to \infty} \frac {4/x - 2}{1 - 4/x}$ = -2

Solving for $\lim \limits_{x\to 2}$ I can factor:

$\lim \limits_{x\to 2}$ f(x) => $\lim \limits_{x\to 2} \frac {2x(2-x)}{(x-2)(x+2)}$ => $\lim \limits_{x\to 2} \frac {-2x}{2-x}$ = -1

My question is: why can't I use the first method to solve for $\lim \limits_{x\to 2}$? It results in 0/0.

Thank you

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    $\begingroup$ As you say, it results in "$0/0$". Better, try $\lim_{y\to0}\frac{4(y+2)-2(y+2)^2}{(y+2)^2-4}$. $\endgroup$ Commented Feb 14, 2018 at 18:54
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    $\begingroup$ It might be worth noting that a $\lim_{x\to \infty}$ can be converted to a (one-sided) limit at a finite point by a substitution $x := \frac{1}{t}$. So for example, $\lim_{x\to \infty} \frac{4x - 2x^2}{x^2 - 4} = \lim_{t\to 0^+} \frac{4 \cdot \frac{1}{t} - 2 (\frac{1}{t})^2}{(\frac{1}{t})^2 - 4}$. $\endgroup$ Commented Feb 14, 2018 at 19:24

3 Answers 3

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$\require{cancel}$

In the first case you could have factored first; it would not have been wrong to have done so:

$$\frac {4x - 2x^2}{x^2 - 4} = = \frac{2x(2-x)}{(x-2)(x+2)} = \frac{-2x(x-2)}{(x-2)(x+2)}$$

Alas, a common factor in the numerator and denominator. Canceling that factor:

$$ = \frac{-2x(\cancel{x-2})}{(\cancel{x-2})(x+2)} = \frac{-2x}{x+2}$$

Now, taking $$\lim_{x\to \infty} \frac{-2x}{x+2},$$ we need to divide the numerator and denominator by $x$ to get:

$$\lim_{x\to \infty} \frac{-2}{1+\frac 2x} = -2.$$

Since factoring and simplifying first, or not doing so, both require the division of numerator and denominator by the largest power of $x$, it is likely more efficient to not bother with simplifying first.


In the second case, as $\lim_{x\to 2}$, factoring and simplifying the function gets immediate results:

$$\lim_{x\to 2} \frac{-2x}{x+2} = -\frac 44 = -1$$

just as you found.


In the second case, using the first approach will require much more work. Let's see what happens:

$$\lim_{x\to 2} \frac {4x - 2x^2}{x^2 - 4}$$

Dividing this (unsimplified) function by the largest power of $x$, which is $x^2$ gives us $$\lim_{x\to 2}\frac{\frac 4x - 2}{1 - \frac 4{x^2}} = \frac {2-2}{1-\frac 44} \to \frac 00$$ which gets us nowhere.

The key about limits $x\to \infty$, e.g. $\lim_{x\to \infty}{\frac 4x} = 0$ annihilates the fraction, whereas $\lim_{x\to 2}{\frac 4x} = 2.$

So the best route for approaching the second limit, is to simplify as much as you can. If factors cancel, all the easier to deal with. If L'hopitals is needed, subsequently, after simplifying, it will be far easier to apply it to the simplified function, than to go back to step one.

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  • $\begingroup$ Thank you. The outcome in part 1 $\lim \limits_{x\to \infty} \frac {-2}{1 + 2/x}$ works both for the Limit to 2 and to Infinity. Why must you factor to solve a finite limit? $\endgroup$ Commented Feb 14, 2018 at 19:28
  • $\begingroup$ I told you that you can factor/simplify the second limit, and then apply the approach you used in (1) to the simplified $$\lim_{x\to 2} \frac{-2x}{x+2}$$ to get (by dividing by $x$: $$\lim_{x\to 2} \frac {-2}{1+\frac 2x} = \frac {-2}{2} = - 1.$$ My only point was, this last step (dividing the simplified function by $x$) is nothing but busy-work, because we can immediately evaluate the simplified function: $$\lim_{x\to 2} \frac{-2x}{x+2} = \frac {-2\cdot 2}{ 2 + 2} = \frac {-4}4 = -1$$ without the need to divide by $x$. $\endgroup$
    – amWhy
    Commented Feb 14, 2018 at 20:00
  • $\begingroup$ Oh, I see - because $\lim \limits_{x\to \infty}$ tends to 0 if x is in the denominator, it is necessary to divide by the largest power in hopes that it removes this problem, whereas when evaluating a finite limit this is not an issue. Thank you for your help. $\endgroup$ Commented Feb 14, 2018 at 21:01
  • $\begingroup$ You're very welcome, user2913891 $\endgroup$
    – amWhy
    Commented Feb 14, 2018 at 21:31
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use that $$\frac{4x-2x^2}{x^2-4}=\frac{2x(2-x)}{(x-2)(x+2)}$$

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If you want to solve by that method you will have to use L'Hospital rule for indeterminate form since $\frac{0}{0} $is indeterminate which means we cannot tell it's value. I think you can solve it by seeing some notes or YouTube video.

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