Find without L'Hospital's rule: $\lim\limits_{x \to 2} \frac{\sqrt{17 - 2x^{2}}\sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} - 9}{(x - 2)^{2}}$

Question

$A = \sqrt{17 - 2x^{2}} \sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} \\ \Rightarrow \lim\limits_{x \to 2} \frac{\sqrt{17 - 2x^{2}}\sqrt[3]{3x^{3} - 2x^{2} + 8x - 5} - 9}{(x - 2)^{2}} \\ = \lim\limits_{x \to 2} \frac{A^{6} - 9^{6}}{(x - 2)^{2}(A^{5} + 9A^{4} + ... + 9^{5})} \\ = \lim\limits_{x \to 2} \frac{(x - 2)^{2}(-72x^{10} - 192x^{9} + 940x^{8} + 2576x^{7} + 874x^{6} + 1992x^{5} - 24543x^{4} - 73908x^{3} - 82540x^{2} - 200414x - 102154)}{(x - 2)^{2}(A^{5} + 9A^{4} + ... + 9^{5})} \\ = \frac{-2.11.3^{10}}{6.9^{5}} = \frac{-11}{3}.$ Am I right? Is there a simple way?

Answer 1

Yes, that works.

You can also use the fact that, if $f(x)=\sqrt{17-2x^2}\sqrt[3]{3x^3-2x^2+8x-5}$, then $f(2)=9$, $f'(2)=0$, and $f''(2)=-\frac{22}3$. So, near $0$,$$f(x)=9-\frac{11}3(x-2)^2+O\bigl((x-2)^3\bigr)$$ and therefore,\begin{align}\lim_{x\to2}\frac{\sqrt{17-2x^2}\sqrt[3]{3x^3-2x^2+8x-5}}{(x-2)^2}&=\lim_{x\to2}\frac{f(x)-f(2)}{(x-2)^2}\\&=-\frac{11}3.\end{align}

Answer 2

The best option is to do a factorization of the polynomials in radicals. The numerator of the expression under limit is $$\sqrt{9-2(x-2)(x+2)}\sqrt[3]{(x-2)(3x^2+4x+16)+27}-9=AB-9$$ where $$A=\sqrt {9-2(x-2)(x+2)}\to 3,B=\sqrt[3]{(x-2)(3x^2+4x+16)+27}\to 3$$ Next note that $$\frac{A-3}{x-2}=\frac{A^2-9}{(A+3)(x-2)}=-\frac{2(x+2)}{A+3}\to - \frac{4}{3}\tag{1}$$ and $$\frac{B-3}{x-2}=\frac{B^3-27}{(x-2)(B^2+3B+9)}=\frac {3x^2+4x+16}{B^2+3B+9}\to \frac{4}{3}\tag{2}$$ Multiplying these limits we see that $$\frac{AB-3A-3B+9}{(x-2)^2}\to-\frac{16}{9}$$ or $$\frac{AB-9}{(x-2)^2}+3\cdot \frac{6-A-B}{(x-2)^2}\to-\frac{16}{9}\tag{3}$$ We now need to figure out how to deal with second term on left above.

Let's observe that $$\frac{A-3}{x-2}+\frac{4}{3}=\frac{3A-(17-4x)}{3(x-2)}=\frac{9A^2-(17-4x)^2}{3(x-2)(3A+(17-4x))}$$ and $$9A^2-(17-4x)^2=-34(x^2-4x+4)=-34(x-2)^2$$ and hence $$\frac{1}{x-2}\left(\frac{A-3}{x-2}+\frac{4}{3}\right)=-\frac{34}{3(3A+(17-4x))}\to-\frac{17}{27}\tag{4}$$ Next we have $$\frac{B-3}{x-2}-\frac{4}{3}=\frac{3B-(4x+1)}{3(x-2)}=\frac {27B^3-(4x+1)^3}{3(x-2)(9B^2+3B(4x+1)+(4x+1)^2)}$$ and $$27B^3-(4x+1)^3=27(3x^3-2x^3+8x-5)-(64x^3+48x^2+12x+1)=17x^3-102x^2+3\cdot 68x-136=17(x^3-6x^2+12x-8)$$ which can be further simplified as $17(x-2)^3$ and hence $$\frac{1}{x-2}\left(\frac{B-3}{x-2}-\frac{4}{3}\right)\to 0\tag{5}$$ Adding $(4)$ and $(5)$ we get $$\frac{A+B-6}{(x-2)^2}\to-\frac{17}{27}$$ and using equation $(3)$ we get $$\frac{AB-9}{(x-2)^2}\to-\frac{16}{9}-\frac{17}{9}=-\frac{11}{3}$$ as desired.