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lim [sqrt(1+3x^4) - sqrt(1-2x) ] / (x+x^2+2x^3) . x->0
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try rationalizing the numerator
\[\lim_{x \rightarrow 0}\frac{\sqrt{1+3x^4}-\sqrt{1-2x}}{x(1+x+2x^2)} \cdot \frac{\sqrt{1+3x^4}+\sqrt{1-2x}}{\sqrt{1+3x^4}+\sqrt{1-2x}}\]
\[\lim_{x \rightarrow 0}\frac{(1+3x^4)-(1-2x)}{x(1+x+2x^2)(\sqrt{1+3x^4}+\sqrt{1-2x})}\]
\[\lim_{x \rightarrow 0}\frac{3x^4+2x}{x(1+x+2x^2)(\sqrt{1+3x^4}+\sqrt{1-2x})}\]
now factor the top and cancel an x
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\[\lim_{x \rightarrow 0}\frac{x(3x^3+2)}{x(1+x+2x^2)(\sqrt{1+3x^4}+\sqrt{1-2x})}\]
\[\lim_{x \rightarrow 0}\frac{3x^3+2}{(1+x+2x^2)(\sqrt{1+3x^4}+\sqrt{1-2x})}\]
what happens if you plug in 0 now?
I'll get my answer . Thank you so much!
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