Not yet. My teacher assigned this problem after we finished the limitView attachment 14317
Have you learned L'Hospital's method?
The only thing i can think of right now it's i can let the cube root be a letter. but I don't know what to do nextIf you haven't you learned L'Hospital's method, then use your knowledge of how the difference of cubes factors.
\(\displaystyle \left( {\frac{{\sqrt[3]{{1 + {\pi ^2}x}}-1}}{x}} \right)\left( {\frac{{{{(\sqrt[3]{{1 + {\pi ^2}x}})}^2} - (\sqrt[3]{{1 + {\pi ^2}x}}) + 1}}{{\sqrt[3]{{1 + {\pi ^2}x}}{)^2} - (\sqrt[3]{{1 + {\pi ^2}x}}) + 1}}} \right) = \frac{{1 + {\pi ^2}x - 1}}{{x\left( {\sqrt[3]{{1 + {\pi ^2}x}}{)^2} - (\sqrt[3]{{1 + {\pi ^2}x}}) + 1} \right)}}\)The only thing i can think of right now it's i can let the cube root be a letter. but I don't know what to do next
how do i get rid of -1 in the numerator in the original formula?\(\displaystyle \left( {\frac{{\sqrt[3]{{1 + {\pi ^2}x}}}}{x}} \right)\left( {\frac{{{{(\sqrt[3]{{1 + {\pi ^2}x}})}^2} - (\sqrt[3]{{1 + {\pi ^2}x}}) + 1}}{{\sqrt[3]{{1 + {\pi ^2}x}}{)^2} - (\sqrt[3]{{1 + {\pi ^2}x}}) + 1}}} \right) = \frac{{1 + {\pi ^2}x - 1}}{{x\left( {\sqrt[3]{{1 + {\pi ^2}x}}{)^2} - (\sqrt[3]{{1 + {\pi ^2}x}}) + 1} \right)}}\)
how do i get rid of -1 in the numerator in the original formula?