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$$\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$$

I've tried using the $A^3-B^3$ identity, but that doesn't help. Also, I tried multiplying every fraction with $\sqrt[3]{A^2}$ to get rid of the roots in the denominator, but that doesn't help either. Can someone suggest a solution? Thanks.

Martin Sleziak
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LearningMath
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4 Answers4

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Let $x-2=t$.

Hence, $$\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )=$$ $$=\lim_{t\rightarrow0^-}\frac{1}{\sqrt[3]{t}}\left(\frac{1}{\sqrt[3]{t+1}}+\frac{1}{\sqrt[3]{t-1}}\right)=$$ $$=\lim_{t\rightarrow0^-}\frac{1}{\sqrt[3]{t+1}\sqrt[3]{t-1}}\lim_{t\rightarrow0^-}\frac{\sqrt[3]{t+1}+\sqrt[3]{t-1}}{\sqrt[3]t}=$$ $$=-\lim_{t\rightarrow0^-}\frac{\sqrt[3]{t+1}+\sqrt[3]{t-1}}{\sqrt[3]t}=$$

$$=-\lim_{t\rightarrow0^-}\frac{2t}{\sqrt[3]t\left(\sqrt[3]{(t+1)^2}-\sqrt[3]{t^2-1}+\sqrt[3]{(t-1)^2}\right)}=$$ $$=-\lim_{t\rightarrow0^-}\frac{2\sqrt[3]{t^2}}{3}=0.$$ I used the following formula $a^3+b^3=(a+b)(a^2-ab+b^2)$,

where $a=\sqrt[3]{t+1}$ and $b=\sqrt[3]{t-1}$.

By this formula we obtain: $$\sqrt[3]{t+1}+\sqrt[3]{t-1}=\frac{t+1+t-1}{\sqrt[3]{t+1)^2}-\sqrt[3]{(t+1)(t-1)}‌​+\sqrt[3]{(t+1)^2}}$$

Michael Rozenberg
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  • I have two questions: Why is there a minus before the limit and how the last expression is zero when we have $\frac{0}{0}$? Thanks. – LearningMath Jul 24 '17 at 14:24
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    @LearningMath The minus happens because $\lim\limits_{t\rightarrow0^-}\sqrt[3]{x+1}\sqrt[3]{x-1}=-1$. The last limit is equal to $0$ because $\frac{t}{\sqrt[3]t}=\sqrt[3]{t^2}\rightarrow0$. – Michael Rozenberg Jul 24 '17 at 14:28
  • How can you substitute the zero in that expression? AFAIK, the product can be written as two separable limits only when they both exist, which is not true for $\frac{1}{\sqrt[3]{t}}$. Can you explain your reasoning here? And what about my second question above. Thank you. – LearningMath Jul 24 '17 at 14:38
  • Thanks for making some clarifications. But I still have trouble understanding the reasoning in this line: $$\lim_{t\rightarrow0^-}\frac{1}{\sqrt[3]{t+1}\sqrt[3]{t-1}}\lim_{t\rightarrow0^-}\frac{\sqrt[3]{t+1}+\sqrt[3]{t-1}}{\sqrt[3]t}$$ How can you split the limit in these two limits, when the second limit evaluates to $\frac{0}{0}$ when $0$ is substituted for $t$? – LearningMath Jul 24 '17 at 14:51
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    @LearningMath The second limit I calculated later, which says this limit exist and I can use the theorem: If exists $\lim\limits_{x\rightarrow a}f(x)$ and exists $\lim\limits_{x\rightarrow a}g(x)$ then exists $\lim\limits_{x\rightarrow a}(f(x)g(x))$ for which $\lim\limits_{x\rightarrow a}(f(x)g(x))=\lim\limits_{x\rightarrow a}f(x)\lim\limits_{x\rightarrow a}g(x)$. – Michael Rozenberg Jul 24 '17 at 16:03
  • Thank you a lot. That was my question. But what if the limit failed to exist? Does that mean that the initial limit doesn't exist or that you should try another approach? – LearningMath Jul 24 '17 at 16:08
  • @LearningMath: for the rule $$\lim_{x\to a} {f(x) g(x)} =\lim_{x\to a} f(x)\cdot \lim_{x\to a} g(x) $$ we only need to ensure that one of the two limits on the right side exists finitely and is non-zero without worrying about the other limit on the right side. For $$\lim_{x\to a}{ f(x) \pm g(x)} =\lim_{x\to a} f(x) \pm\lim_{x\to a} g(x) $$ we only need to ensure that one of the two limits on right side exists finitely. – Paramanand Singh Jul 24 '17 at 16:50
  • @ParamanandSingh Why is that? And why for the product the restriction non-zero is needed? Thanks? – LearningMath Jul 24 '17 at 16:52
  • @ParamanandSingh And for the Michael Rozenberg last comment: why I can't split the limit in two limits where both evaluate to indeterminate forms and the solve for both? – LearningMath Jul 24 '17 at 16:56
  • @LearningMath: I am glad you noticed the non-zero thing for product rule. It is absolutely necessary for product rule that one of the limits is non-zero if you want to split without worrying about the other limit. On the other hand if you already have information about both limits then it is not a problem. While evaluating limits in step by step manner you can not find limit of all terms and factors at once and you need to do split to handle one limit at a time without worrying about other terms and factors. This is where the rules given in my comment help. – Paramanand Singh Jul 24 '17 at 16:56
  • @LearningMath: also you should not try to manipulate your expression in such a form that you are unable to easily evaluate both the limits. The expression should always be manipulated so that one of the limits can be easily evaluated and the product /sum/difference rule can be applied. If such manipulation is not possible then these rules don't work and you need Taylor or L'Hospital's Rule. – Paramanand Singh Jul 24 '17 at 17:00
  • @I still don't understand why it is necessary for it to be non-zero. For example, the one limit can be 0, the other can be any value L, and still the limit we started with will be zero. Where is the problem? Also, what happens if we find one of the limits not to exist? We try another approach or we can conclude that the initial limit doesn't exist? – LearningMath Jul 24 '17 at 17:00
  • @LearningMath: Most textbooks don't present the rules in the manner I have written them, but almost always the books use these rules I examples of limit problems. These are not difficult to prove, but then I am not sure if you are aware of a typical proof of limit laws based on $\epsilon, \delta$. – Paramanand Singh Jul 24 '17 at 17:02
  • @LearningMath If one of two limits does not exist we can not use this theorem. But in our case two limits are exist and we are done! I really don't understand what are you looking for? – Michael Rozenberg Jul 24 '17 at 17:03
  • @ParamanandSingh "The expression should always be manipulated so that one of the limits can be easily evaluated and the product /sum/difference rule can be applied. If such manipulation is not possible then these rules don't work and you need Taylor or L'Hospital's Rule." Why is this so? – LearningMath Jul 24 '17 at 17:04
  • @LearningMath: when you have information about both the limits then you don't care about zero versus non-zero. But many times it happens that you have information only about one of the limits (like it is easier to evaluate) then you need it to be non-zero to make a split without having any information about the other limit. – Paramanand Singh Jul 24 '17 at 17:06
  • That both limits exist is an information you get at the end. The rules are sufficient to do split in the intermediate steps without knowing what happens at the end. – Paramanand Singh Jul 24 '17 at 17:08
  • @ParamanandSingh And why you should care about the first limit to be non-zero when you don't have info about the second one when you should evaluate it at the end anyway. How is the first limit being non-zero relevant to this? For example, you split the limit, you evaluate the first and it is zero, then you go on to evaluate the second: if it exists, you have a value L which is sum of the first and the second limit, if not, try another apporach. – LearningMath Jul 24 '17 at 18:22
  • @LearningMath: try to make a split like $$\lim_{x\to 0}x(1/x)=\lim_{x\to 0}x\cdot \lim_{x\to 0}1/x$$ and the first limit on right is $0$. Then you have a problem you can't do anything now with $0\cdot\lim_{x\to 0}(1/x)$. – Paramanand Singh Jul 24 '17 at 21:27
  • @ParamanandSingh Why you can't do anything now? Yes, I agree, in this case you can't, but what if the second limit evaluates to a value? Then the original limit is zero. – LearningMath Jul 24 '17 at 21:30
  • Let us continue this discussion in chat. – Paramanand Singh Jul 24 '17 at 21:32
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An interesting discussion ensued between myself and OP in comments to one of the answers here about the use of algebra of limits (rules relating to limit of sum, difference, product and quotient of two functions). It would be best to put all of that with some more details into an answer here for the benefit of everyone.

First the answer to the current question. We can proceed as follows: \begin{align} L&=\lim_{x\to 2^{-}}\frac{1}{\sqrt[3]{x^{2}-3x+2}}+\frac{1}{\sqrt[3]{x^{2}-5x+6}}\notag\\ &= \lim_{x\to 2^{-}}\frac{1}{\sqrt[3]{(x-2)(x-1)}}+\frac{1}{\sqrt[3]{(x-2)(x-3)}}\notag\\ &= \lim_{x\to 2^{-}}\frac{\sqrt[3]{x-1}+\sqrt[3]{x-3}}{\sqrt[3]{(x-1)(x-2)(x-3)}}\notag\\ &= \lim_{x\to 2^{-}}\frac{1}{\sqrt[3]{(x-1)(x-3)}}\cdot\frac{\sqrt[3]{x-1}+\sqrt[3]{x-3}}{\sqrt[3]{x-2}} \notag\\ &= \lim_{x\to 2^{-}}\frac{1}{\sqrt[3]{(x-1)(x-3)}}\cdot\lim_{x\to 2^{-}}\frac{\sqrt[3]{x-1}+\sqrt[3]{x-3}}{\sqrt[3]{x-2}} \tag{1}\\ &=(-1)\cdot\lim_{x\to 2^{-}}\frac{2x-4}{\sqrt[3]{x-2}\{\sqrt[3]{(x-1)^{2}}-\sqrt[3]{(x-1)(x-3)}+\sqrt[3]{(x-3)^{2}}\}}\notag\\ &=-\lim_{x\to 2^{-}}\frac{1}{\sqrt[3]{(x-1)^{2}}-\sqrt[3]{(x-1)(x-3)}+\sqrt[3]{(x-3)^{2}}}\cdot\lim_{x\to 2^{-}}2(x-2)^{2/3}\tag{2}\\ &=-\frac{1}{3}\cdot 0=0\notag \end{align}

The algebra of limits has been used at two places here which are marked using equation numbers $(1),(2)$. The limits laws are normally stated in textbooks as follows:

If both the limits $\lim_{x\to a} f(x) $ and $\lim_{x\to a} g(x) $ exist then so does $\lim_{x\to a} (f(x) \circ g(x)) $ and $$\lim_{x\to a} (f(x)\circ g(x)) =\lim_{x\to a} f(x) \circ \lim_{x\to a} g(x) \tag{3}$$ where $\circ$ is one of the usual binary operations $+, -, \times, /$ and in the specific case of $\circ=/$ the limit $\lim_{x\to a} g(x)$ must be non-zero.

However in most common applications of these laws we have information only about one of the limits on the right of $(3)$ at a time and the information about the other limit is available later. The rules above can be generalized for this scenario by just requiring that in case of $+, - $ only one of the limits on right of $(3)$ exists finitely and in case of $\times, /$ only one of the limits on right of $(3)$ exists finitely and is non-zero.

And then the equation $(3)$ has to be interpreted in a slightly more general manner. To be specific let one of the limits on the right be $\lim_{x\to a} f(x) $ whose existence is guaranteed. Then the limiting behavior of $f(x) \circ g(x) $ is same as that of $g(x) $ in the sense that:

  • if $g(x) $ diverges (to $\pm\infty$) as $x\to a$ then so does $f(x) \circ g(x) $.
  • if $g(x) $ oscillates (finitely or infinitely) as $x\to a$ then so does $f(x) \circ g(x) $.
  • if $g(x) $ tends to a finite limit as $x\to a$ then so does $f(x) \circ g(x) $ and it's limit is given by equation $(3)$.

One can notice that in this manner based on information about limit of $f(x) $ we can effectively eliminate the dependency of limit of $f(x) \circ g(x) $ on the limit of $f(x) $ and continue to focus on $g(x) $. Thus in our solution above in equation $(1)$ we see that the first limit is $(-1)$ and we can now work on the second limit after this step without ever having to come back to this step. In equation $(2)$ the situation is simpler as we have information about both the limits and we apply limit laws in their usual form.

You can have a look at the use of algebra of limits for the evaluation of a slightly more complicated limit in this answer.

Paramanand Singh
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  • Why is the restriction of one of the limits to be non-zero imposed? Here is an example: $$\lim_{x \to 0}x \frac{\sin x}{x} =\lim_{x \to 0}x \cdot \lim_{x \to 0}\frac{\sin x}{x}=0\cdot 1=0$$ We suppose here that we don't know yet how to calculate $ \lim_{x \to 0}\frac{\sin x}{x}$ and first we get $\frac{1}{0}$ when we split the limit. So, in this case, we have one of them to be zero but still we can split it and calculate the second limit later to obtain a finite value $1$ for it. So, why the non-zero restriction is needed? – LearningMath Jul 25 '17 at 19:58
  • @LearningMath: Suppose that one of the limits ie $\lim_{x\to a} f(x) =0$ and we are dealing with product rule. Then in this case only the last of my 3 bullet points remains valid. The first two points do not hold true. And this means means that the rule is usable only if the limit of other factor ie $g(x) $ is also known to exist finitely. Thus the decision of split can not be made without the knowledge about $g(x) $. In your example limit of $g(x) =(\sin x) /x$ exists and that's why it worked. You could not do anything if $g(x) =1/x$ or $g(x) =\sin(1/x)$. Cont'd.. – Paramanand Singh Jul 25 '17 at 20:16
  • @LearningMath: if $\lim_{x\to a} f(x)$ is non-zero then we have all 3 bullet points valid and we have a well defined conclusion for all kinds of $g(x) $. – Paramanand Singh Jul 25 '17 at 20:18
  • Yes, I am aware of the bullet points. But, my point is, we should not restrict ourselves to not do this, because if we find that the second limit exists then the question is solved easily. If not, we go back. Before the splitting we assume that the second limit exists and proceed to evaluate it. If we get a value, the question is solved, if not we get a contradiction, so we go back and try another technique. – LearningMath Jul 25 '17 at 20:22
  • @LearningMath: to express differently, the limiting behavior of $f(x) g(x) $ can be deduced from that of $g(x) $ only if limit of $f(x) $ is non-zero. If the limit of $f(x) $ is zero then the limiting behavior of $f(x) g(x) $ can not be deduced from that of $g(x) $ alone and we need to analyze the whole product together rather than in isolation. Thus the split is not possible. – Paramanand Singh Jul 25 '17 at 20:22
  • @LearningMath: the rules precisely are designed to avoid looking back. The decision can be made at each step without knowing what happens next. Looking back and trying once again is just wasting time. – Paramanand Singh Jul 25 '17 at 20:25
  • That's ok but I am asking whether what I say is mathematically valid. To suppose the existence of the second limit, then proceed to evaluate it and reach to a solution or a contradiction. – LearningMath Jul 25 '17 at 20:27
  • @LearningMath: you can proceed as you say provided you are willing to take the pain to come back in case of contradiction. And then there is no point of split. When you do the split you actually replace the expression $\lim_{x\to a} f(x) $ by its limit $L$ and forget about the function $f$. All that remains after split is the limit $L$ and not $f(x) $. If you want to come back in case of contradiction then you have to keep on using $\lim_{x\to a} f(x) $ and effectively there is no split. – Paramanand Singh Jul 25 '17 at 20:31
  • Thanks. I think, this should be mentioned in your already comprehensive answer to be more complete on the matter. Thank you for the fruitful discussion, now I understand it more thoroughly. – LearningMath Jul 25 '17 at 20:34
  • I am accepting Michael's answer because of the upvotes. Both answers are really, really great, but I have to accept only one. – LearningMath Jul 25 '17 at 20:36
  • @LearningMath: You may also look at this answer https://math.stackexchange.com/a/1783818/72031 which describes the same thing but in a more important context. – Paramanand Singh Jul 25 '17 at 20:36
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My approach begins with $t=x-2$ as did Michael's, but then we note $(1+t)^{-1/3}-(1-t)^{-1/3}\in\mathcal{O}(t)$ for small $t$ by the binomial theorem. Multiplying by $t^{-1/3}$ gives a positive power of $t$, with limit $0$ as in his answer.

J.G.
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Note that \begin{align} x^2-3x+2&=(2-x)(1-x)=(2-x)(2-x-1) \\[4px] x^2-5x+6&=(2-x)(3-x)=(2-x)(2-x+1) \end{align} Set $2-x=t^3$, so your expression becomes $$ \frac{1}{\sqrt[3]{t^3(t^3-1)}}+\frac{1}{\sqrt[3]{t^3(t^3+1)}} =\frac{1}{t}\frac{\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}}{\sqrt[3]{t^6-1}} $$ and the limit is $$ \lim_{t\to0^+}\frac{1}{\sqrt[3]{t^6-1}}\frac{\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}}{t} $$ The first fraction has limit $-1$, whereas the second fraction has limit $f'(0)$, where $f(t)=\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}$. Since $$ f'(t)=3t^2(t^3+1)^{-2/3}+3t^2(t^3-1)^{-2/3} $$ we have $f'(0)=0$.

Alternatively, \begin{align} \frac{1}{\sqrt[3]{t^3(t^3-1)}}+\frac{1}{\sqrt[3]{t^3(t^3+1)}} &=\frac{(1+t^3)^{-1/3}-(1-t^3)^{-1/3}}{t}\\[6px] &=\frac{1-\frac{1}{3}t^3-1-\frac{1}{3}t^3+o(t^3)}{t}\\[6px] &=-\frac{2}{3}t^2+o(t^2) \end{align}

egreg
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