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

Question

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

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.

Answer 1

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

Answer 2

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.

Answer 3

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}

Answer 4

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.

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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}

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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.

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