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$$ \lim_{x \to 1} \left( \frac{2}{1-x^2} - \frac{3}{1-x^3} \right)$$

In my opinion the function is not defined at $ x = 1 $ but somehow when I look at the graph, it's continuous and there is no break. I learned to look for points where my function is not defined due to division by zero.

So my example here is: I should look for the limit as $x \rightarrow 1$ but I don't know how to do this. I'm not allowed to use L'Hospital. I know how to look for limits if my variable ($n$ or $x$) "runs" to infinity. I know that $\frac{1}{n}$ when $n \to \infty$ is $0$. I just "know" that. And if $n$ would "go" or "run" or "tend" (what is the right way to call it) to $1$, the limit would be $1$ just by "imagining the $n$ as $1$". Or is that the wrong way? I'm still learning, so please let me know the right way to do it.

Back to the example. Is there an approach of calculating the limit? In my book they suggest to replace $x$ with a sequence $ x_n$ and $ x_n \rightarrow 1$ and then let $ n \rightarrow \infty $

Lord_Farin
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loop
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4 Answers4

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Hint: expand/simplify the expression to get $-\dfrac{2x+1}{(x+1)(x^2+x+1)}$.

Amzoti
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  • how? really? I totally get that expansion. But how would i calculate the limit then? – loop Sep 13 '13 at 14:39
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    just plug in 1 everywhere you see $x$ – J. W. Perry Sep 13 '13 at 14:40
  • @DominiqueLüber as J. W. Perry suggested, just plug in $x=1$. Also, to clarify, the suggestion provided by book works only if you know the limit exists (and simplifying the expression shows that the limit exists). –  Sep 13 '13 at 14:44
  • yeah but if i plug in x=1 then my whole function turns to zero? I'm trying to simplify and expand right now .. but somehow I made a mistake i guess. got -6 as a result. – loop Sep 13 '13 at 14:46
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    @DominiqueLüber wherever you see an $x$ in the simplified expression, replace it with 1. You should get $- \frac{1}{2}$. By the way, I can show you how I got the simplified expression, if you need. –  Sep 13 '13 at 14:48
  • i add the (1-x^2) and (1-x^3) to both expressions right? then i have $ \frac{2(1-x^3) - 3(1-x^2)}{(1-x^2)(1-x^3)} $ – loop Sep 13 '13 at 14:52
  • at the end i take out $ x^3 $? I think I dont have made an error now but somehow I will get 0 if I plug in 1 – loop Sep 13 '13 at 14:52
  • @DominiqueLüber yes, once you have $\frac{2(1-x^3) - 3(1-x^2)}{(1-x^2)(1-x^3)}$, don't take out $x^3$, BUT use that $1-x^2=(1-x)(1+x)$ and $1-x^3=(1-x)(1+x+x^2)$. Now you can get rid of the $(x-1)$ in the denominator and numerator! –  Sep 13 '13 at 14:56
  • let us continue this discussion in chat –  Sep 13 '13 at 14:57
  • how would anybody expect me to come up with that? thats frustrating. i mean, yes. of course i know that $(1-x^2)$ is $(1-x)(1+x)$ but to see this, you need to see 324825242342342 options everytime you have an example. I feel really bad – loop Sep 13 '13 at 14:57
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Hint: Consider the sequence $\{x_n\}$ defined by $x_n = 1 + \frac{1}{n}$. What is $\frac{2}{1 - x_n^2} - \frac{3}{1 - x_n^3}$? Can we conclude that $$ \lim_{x \to 1} \frac{2}{1 - x^2} - \frac{3}{1 - x^3} = \lim_{n \to \infty} \frac{2}{1 - x_n^2} - \frac{3}{1 - x_n^3}? $$

tylerc0816
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    We can conclude that if the limit on the right-hand side is $L$, then if the limit on the left-hand side exists, it must also be $L$. But, this leaves open showing that the limit on the left-hand side exists. [In fact, this method is often used to show that a limit doesn't exist: we find a sequence which results in limit $A$ and another sequence which results in limit $B \neq A$.] – Rebecca J. Stones Sep 13 '13 at 14:35
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The only problem with your book's hint of using a specific sequence $x_n$ that approaches $1$ is that this only shows what the limit must be if you already know that the limit exists. But it's possible that the limit doesn't exist, and that you'll get one answer for one sequence $x_n \to 1$, but get a different answer for another sequence $y_n \to 1$ (for example if $x_n < 1$ and $y_n > 1$). Thus, when you are asked to evaluate a limit, it's a good idea to confirm whether you are allowed to assume the limit exists. If not, then the standard $(\delta, \epsilon)$ approach is a good way to go, once you learn that, or also using algebraic manipulations etc. like another answer posted.

user2566092
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Consider the expression $$\lim_{x \to 1} \left( \frac1{1-x} - \frac1{1-x} \right) $$

This is "obviously" zero, even though both terms blow up as $x \to 1$, because the singularities cancel.

The same thing happens here.

The basic identity needed here is $1-x^n =(1-x)(1+x+x^2+...+x^{n-1}) $.

Using this,

$\begin{align} \frac{2}{1-x^2} - \frac{3}{1-x^3} &=\frac{2}{(1-x)(1+x)} - \frac{3}{(1-x)(1+x+x^2)}\\ &=\frac1{1-x}\left(\frac{2}{1+x} - \frac{3}{1+x+x^2}\right)\\ &=\frac1{1-x}\left(\frac{2(1+x+x^2)-3(1+x)}{(1+x)(1+x+x^2)}\right)\\ &=\frac1{1-x}\left(\frac{-1-x+2x^2}{(1+x)(1+x+x^2)}\right)\\ &=\frac1{1-x}\left(\frac{(2x+1)(x-1)}{(1+x)(1+x+x^2)}\right)\\ &=\frac{(2x+1)(x-1)}{(1-x)(1+x)(1+x+x^2)}\\ &=\frac{-(2x+1)}{(1+x)(1+x+x^2)} \quad \text{ (except where } x=1)\\ \end{align} $

This final expression has no problem being evaluated as $x \to 1$. Its value is $\frac{-3}{2\cdot 3} =-\frac12 $.

Another way to do this is to let $x = 1+y$, so $y \to 0$ is the same as $x \to 1$. I find it easier to see a limit with a variable going to $0$, so I make this kind of transformation whenever possible.

Since $1-x^2 = 1-(1-y)^2 =1-(1-2y+y^2) =2y-y^2 $ and $1-x^3 = 1-(1-y)^3 =1-(1-3y+3y^2-y^3) =3y-3y^2+y^3 $,

$\begin{align} \frac{2}{1-x^2} - \frac{3}{1-x^3} &=\frac{2}{2y-y^2}- \frac{3}{3y-3y^2+y^3}\\ &=\frac1{y}\left(\frac{2}{2-y}- \frac{3}{3-3y+y^2}\right) \quad \text{ (factoring }y\text{ from each denominator)}\\ &=\frac1{y}\frac{2(3-3y+y^2)-3(2-y)}{(2-y)(3-3y+y^2)}\\ &=\frac1{y}\frac{-3y+2y^2}{(2-y)(3-3y+y^2)}\\ &=\frac{-3+2y}{(2-y)(3-3y+y^2)}\\ \end{align} $

Again, this has no problem as $y \to 0$, and gives $-\frac12$ as before (as it should).

Note that, by doing it as $y \to 0$, we do not need to know the factorizations of $1-x^2$ and $1-x^3$. The fact that $1-x$ divides these becomes the more obvious fact that $y$ divides the expressions for $1-(1-y)^2$ and $1-(1-y)^3$.

marty cohen
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