a_ng116
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1) \lim_{x\rightarrow 0}\frac {\sqrt[3]{1+x^2} -1}{x^2}
I tried doing a difference of cubes to the top and I got: \frac{\sqrt[3]{1+x^2}-1 ((\sqrt[3]{1+x^2})^2 + \sqrt[3]{1+x^2} + 1)} {x^2}
I know you need to get rid of the x^2 on the bottom of the equation but now I'm stuck. Am I approaching this the right way and does anyone have any suggestions? Please and thank you.
I tried doing a difference of cubes to the top and I got: \frac{\sqrt[3]{1+x^2}-1 ((\sqrt[3]{1+x^2})^2 + \sqrt[3]{1+x^2} + 1)} {x^2}
I know you need to get rid of the x^2 on the bottom of the equation but now I'm stuck. Am I approaching this the right way and does anyone have any suggestions? Please and thank you.
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