Question posed by my peruvian teacher of calculus

Factoring it out, we have:

$\displaystyle L=\lim_{x \rightarrow {1}} \ \ \displaystyle{\sqrt[3]{x} \cdot \frac{1-\sqrt[6]{x}}{1-x}}$

With variable substitution, comes:

$\displaystyle t=\sqrt[6]{x} \implies L=\lim_{t \rightarrow {1}} \ \ \displaystyle{t^{2} \cdot \frac{1-t}{1-t^{6}}}$

From the sum of terms on a geometric progression, we know:

$\displaystyle {\frac{1-t^{6}}{1-t}=1+t+t^{2}+t^{3}+t^{5}+t^{6}}$

Thus, we have:

$\displaystyle L=\lim_{t \rightarrow {1}} \ \ \displaystyle{\frac{t^{2}}{1+t+t^{2}+t^{3}+t^{5}+t^{6}}}$

Utilizing the limits' property that $\displaystyle L=\lim_{x \rightarrow {x_{0}}} \ \ \displaystyle{\frac{f(x)}{g(x)}}=\displaystyle{\frac{f(x_{0})}{g(x_{0})}}$ , when both functions are polynomials, we get:

$\displaystyle L=\lim_{t \rightarrow {1}} \ \ \displaystyle{\frac{t^{2}}{1+t+t^{2}+t^{3}+t^{5}+t^{6}}}={\frac{1^{2}}{1+1+1^{2}+1^{3}+1^{5}+1^{6}}}=\frac{1}{6}$

Hence, $a+b=\boxed{7}$ .

Let $x = y^6$ , then:

\(\begin{array} {} L & = \displaystyle \lim_{x\rightarrow 1} \dfrac {\sqrt [3] {x} - \sqrt{x}}{1-x} = \displaystyle \lim_{y\rightarrow 1} \dfrac {y^2 - y^3}{1-y^6} \\ & = \displaystyle \lim_{y\rightarrow 1} \dfrac {y^2(1 - y)}{(1-y)(1+y+y^2+y^3+y^4+y^5)} \\ & = \displaystyle \lim_{y\rightarrow 1} \dfrac {y^2}{1+y+y^2+y^3+y^4+y^5} = \frac{1}{6} \end{array} \)

$\Rightarrow a + b = 1+6 = \boxed {7}$

As the (Numerator & Denominator) $\rightarrow 0$ as $x \rightarrow 1$ , we can use L-Hopital's Rule. Using the L-Hopital Rule, we get

$L=\lim_{x\to 1}\frac{x^{\frac{1}{3}}-x^{\frac{1}{2}}}{1-x}$

$=\lim_{x\to 1}\frac{\frac{x^{\frac{-2}{3}}}{3}-\frac{x^{\frac{1}{2}}}{2}}{-1} =\frac{1}{6}=\frac{a}{b}$

$\Rightarrow a+b = 1+6=\boxed{7}$

P.S.

L- Hopital's Rule at a glance: In its simplest form, L'Hôpital's rule states that for functions f and g which are differentiable on an open interval I except possibly at a point c contained in I:

If

$\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty$ , and $\lim_{x\to c}\frac{f'(x)}{g'(x)}$ exists, and $g'(x)\neq 0$ for all x in I with x ≠ c, then

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}$ .

Source: Wikipedia. Thanks for patiently observing.