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

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

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

• i there a plus sign in the middle? – Dr. Sonnhard Graubner Jul 24 '17 at 14:11
• @Dr.SonnhardGraubner If you mean between the two fractions, then yes. – LearningMath Jul 24 '17 at 14:12
• The limit is $\infty+\infty=\infty$ – Raffaele Jul 24 '17 at 16:05

Let $x-2=t$.

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

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

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

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

• 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
• @LearningMath The minus happens because $\lim\limits_{t\rightarrow0^-}\sqrt{x+1}\sqrt{x-1}=-1$. The last limit is equal to $0$ because $\frac{t}{\sqrtt}=\sqrt{t^2}\rightarrow0$. – 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 $\frac{1}{\sqrt{t}}$. 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: $$\lim_{t\rightarrow0^-}\frac{1}{\sqrt{t+1}\sqrt{t-1}}\lim_{t\rightarrow0^-}\frac{\sqrt{t+1}+\sqrt{t-1}}{\sqrtt}$$ 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
• @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

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{x^{2}-3x+2}}+\frac{1}{\sqrt{x^{2}-5x+6}}\notag\\ &= \lim_{x\to 2^{-}}\frac{1}{\sqrt{(x-2)(x-1)}}+\frac{1}{\sqrt{(x-2)(x-3)}}\notag\\ &= \lim_{x\to 2^{-}}\frac{\sqrt{x-1}+\sqrt{x-3}}{\sqrt{(x-1)(x-2)(x-3)}}\notag\\ &= \lim_{x\to 2^{-}}\frac{1}{\sqrt{(x-1)(x-3)}}\cdot\frac{\sqrt{x-1}+\sqrt{x-3}}{\sqrt{x-2}} \notag\\ &= \lim_{x\to 2^{-}}\frac{1}{\sqrt{(x-1)(x-3)}}\cdot\lim_{x\to 2^{-}}\frac{\sqrt{x-1}+\sqrt{x-3}}{\sqrt{x-2}} \tag{1}\\ &=(-1)\cdot\lim_{x\to 2^{-}}\frac{2x-4}{\sqrt{x-2}\{\sqrt{(x-1)^{2}}-\sqrt{(x-1)(x-3)}+\sqrt{(x-3)^{2}}\}}\notag\\ &=-\lim_{x\to 2^{-}}\frac{1}{\sqrt{(x-1)^{2}}-\sqrt{(x-1)(x-3)}+\sqrt{(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.

• 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

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.

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 $$\frac{1}{\sqrt{t^3(t^3-1)}}+\frac{1}{\sqrt{t^3(t^3+1)}} =\frac{1}{t}\frac{\sqrt{t^3+1}+\sqrt{t^3-1}}{\sqrt{t^6-1}}$$ and the limit is $$\lim_{t\to0^+}\frac{1}{\sqrt{t^6-1}}\frac{\sqrt{t^3+1}+\sqrt{t^3-1}}{t}$$ The first fraction has limit $-1$, whereas the second fraction has limit $f'(0)$, where $f(t)=\sqrt{t^3+1}+\sqrt{t^3-1}$. 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{t^3(t^3-1)}}+\frac{1}{\sqrt{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}