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Calculate: $$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$

The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).

mathp
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5 Answers5

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The conjugation/rationalization approach works, but is somewhat tedious. The key is to introduce the right terms in order to force the numerator and denominator into a differences of integer powers:

$$\begin{align} \frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}&=\frac{\left((19-x)^2\right)^{\frac14}-\left(2^4(13+x)\right)^{\frac14}}{(11-x)^{\frac13}-\left((x-1)^3\right)^{\frac13}}\\[1ex] &=\frac{a^{\frac14}-b^{\frac14}}{c^{\frac13}-d^{\frac13}}\\[1ex] &=\frac{a^{\frac14}-b^{\frac14}}{c^{\frac13}-d^{\frac13}}\times\frac{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}\\[1ex] &=\frac{\left(a^{\frac14}\right)^4-\left(b^\frac14\right)^4}{\left(c^{\frac13}\right)^3-\left(d^\frac13\right)^3}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}\\[1ex] &=\frac{a-b}{c-d}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}} \end{align}$$

(where I hope the replacements of $a,b,c,d$ are obvious)

Upon simplification (with $x\neq3)$, we have

$$\frac{a-b}{c-d}=\frac{153-54x+x^2}{12-4x+3x^2-x^3}=\frac{(51-x)(3-x)}{(3-x)(4+x^2)}=\frac{51-x}{4+x^2}$$

and the remaining fraction of rational powers is continuous at $x=3$. Then the limit is

$$\lim_{x\to3}\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}=\frac{48}{13}\times\frac{12}{256}=\boxed{\frac9{52}}$$

user170231
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  • Could you walk me through the process of transformations with a's, b's, c's and d's (if you know what I mean)? IN the first step, you want to have all elements in nominator and denominator, respectively, in the same powers, that's why you "push higher powers down" to pack them in lower powers. Then, once they are unified, you make them a difference of intiger powers. My 2 questions are following. Do you know for sure that you need to start from powers that are unified "downwards"? Second - why at the start are you sure that once you have positive intigers you will be able to eliminate roots? – mathp Jan 29 '21 at 21:50
  • I like your solution very much, I was just wondering how much is it scalable for other similar cases. – mathp Jan 29 '21 at 21:51
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    This approach relies on the well-known result,$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-1}+b^n)$$For an expression like $a^{\frac1\mu}-b^{\frac1\nu}$, you have to choose not just the larger of $\mu$ and $\nu$, but their LCM, to take the role of $n$, because otherwise there is no reduction to $a-b$ after you introduce the extra terms. See for yourself: what happens if you try to rewrite the numerator as a difference of squares (to eliminate the square root) instead of $4$th powers? What happens if you change the $4$th root for a cubic root (noting that $2$ and $3$ are coprime)? – user170231 Jan 29 '21 at 22:45
  • As to your second question, we know cancellation will take place because both numerator and denominator in the starting expression vanish at $x=3$. We rationalize both of them in order to make them polynomials that have a root at $x=3$, which can then be factorized. – user170231 Jan 29 '21 at 22:54
  • @mathp: You may want to learn the generalization of this technique. – user21820 Feb 07 '21 at 05:21
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Let $y=x-3$.

Then it's $\lim\limits_{y\to0}\dfrac{\sqrt{16-y}-2\sqrt[4]{16+y}}{\sqrt[3]{8-y}-y-2}.$

Using binomial series, that's $\dfrac{(4-\frac18y\dots)-2(2+\frac1{32}y\dots)}{(2-\frac1{12}y...)-y-2}=\dfrac{-3/16}{-13/12}=\dfrac9{52.}$

J. W. Tanner
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HINT.-Use the identity $A^3-B^3=(A-B)(A^2+AB+B^2)$ with $A=\sqrt[3]{11-x}$ and $B=x-1$ and multiplying numerator and denominator by $(A^2+AB+B^2)$ you' ll get again the undeterminated form $\dfrac{(4-4)(4+4+4)}{8-8}$. But the new denominator being equal to $11-x-(x-1)^3$ you have its derivative is equal to $-13\ne0$ so the undetermination disappear after apply Hospital's rule.

Piquito
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$$\frac{\sqrt{19-x}-2 \sqrt[4]{x+13}}{\sqrt[3]{11-x}-x+1}$$ plug $x=y+3$ so we get $$\underset{y\to 0}{\text{lim}}\frac{\sqrt{16-y}-2 \sqrt[4]{y+16}}{\sqrt[3]{8-y}-y-2}$$ we have the following expansions as $y\to 0$ $$ \begin{array}{rll} \sqrt{16-y} &\sim &4-\frac{y}{8} \\ 2 \sqrt[4]{y+16} &\sim& \frac{y}{16}+4 \\ \sqrt[3]{8-y} &\sim& 2-\frac{y}{12} \\ \end{array} $$ So the limit can be written as $$\underset{y\to 0}{\text{lim}}\frac{4-\frac{y}{8}-\left(\frac{y}{16}+4\right)}{2-\frac{y}{12}-y-2}=\lim_{y\to0}\frac{-\frac{3}{16}y}{-\frac{13}{12}y}=\frac{9}{52}$$

Raffaele
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Here is a simple strategy. Find the limits of those radicals and subtract that limit from them.

Thus for numerator the term $\sqrt{19-x}\to 4$ and hence subtract and add $4$ to get $$\sqrt{16-x}-4+4$$ The other term in numerator is $2\sqrt[4]{13+x}$ which tends to $4$ and replace it with $$2\sqrt[4]{13+x}-4+4$$ and see that that those extra $4$'s added in both terms cancel out (well such limit problems are designed in such manner that numerator and denominator tend to $0$ and thus we have the desired cancelation here). Thus numerator equals $$\sqrt{19-x}-4-(2\sqrt[4]{13+x}-4)$$ and similarly denominator equals $$\sqrt[3]{11-x}-2-(x-3)$$ Now put $x-3=h$ so that $h\to 0$ and divide each term in numerator and denominator by $h$ to get $$\lim_{h\to 0}\dfrac{\dfrac{\sqrt{16-h}-4}{h}-2\cdot\dfrac{\sqrt[4]{16+h}-2}{h}}{\dfrac{\sqrt[3]{8-h}-2}{h}-1}$$ You can now evaluate the limit of three fractions easily and get the final answer as $$\frac{(-1/8)-2(1/32)} {(-1/12)-1} =\frac{9}{52}$$

Paramanand Singh
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