# Use Taylor's theorem to evaluate the following limits. lim_{xrightarrow0}frac{3sin^2(x)+2sin^4(x)}{3xtan(x)}

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Evaluate limit using Taylor’s theorem.
Given:
$\underset{x\to 0}{lim}\frac{3{\mathrm{sin}}^{2}\left(x\right)+2{\mathrm{sin}}^{4}\left(x\right)}{3x\mathrm{tan}\left(x\right)}$
Taylor series expansion of trigonometric functions,
${\mathrm{sin}}^{2}x={x}^{2}-\frac{{x}^{4}}{3}+\frac{2{x}^{6}}{45}-\frac{{x}^{8}}{315}+...$
${\mathrm{sin}}^{4}x={x}^{4}-\frac{2{x}^{6}}{3}+\frac{{x}^{8}}{5}-\frac{34{x}^{10}}{945}+...$
$\mathrm{tan}x=x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\frac{17{x}^{7}}{315}+\frac{62{x}^{9}}{2835}+...$
Substitute the series,
$\underset{x\to 0}{lim}\frac{3\left({x}^{2}-\frac{{x}^{4}}{3}+\frac{2{x}^{6}}{45}-\frac{{x}^{8}}{315}+...\right)+2\left({x}^{4}-\frac{2{x}^{6}}{3}+\frac{{x}^{8}}{5}-\frac{34{x}^{10}}{945}+...\right)}{3x\left(x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\frac{17{x}^{7}}{315}+\frac{62{x}^{9}}{2835}+...\right)}$
$\underset{x\to 0}{lim}\frac{\left(3{x}^{2}-\frac{3{x}^{4}}{3}+\frac{3\left(2{x}^{6}\right)}{45}-\frac{3{x}^{8}}{315}+...\right)+\left(2{x}^{4}-\frac{2\left(2{x}^{6}\right)}{3}+\frac{2{x}^{8}}{5}-\frac{2\left(34{x}^{10}\right)}{945}+...\right)}{\left(3{x}^{2}+\frac{3{x}^{4}}{3}+\frac{3\cdot 2{x}^{6}}{15}+\frac{3\cdot 17{x}^{8}}{315}+\frac{3\cdot 62{x}^{10}}{2835}+...\right)\right)}$
Divide ${x}^{2}$ in both numerator and denominator,
$\underset{x->0}{lim}\frac{\left(3-\frac{3{x}^{2}}{3}+\frac{3\left(2{x}^{4}\right)}{45}-\frac{3\left({x}^{6}\right)}{315}+...\right)+\left(2{x}^{2}-\frac{2\left(2{x}^{4}\right)}{3}+\frac{2{x}^{6}}{5}-\frac{2\left(34{x}^{8}\right)}{945}+...\right)}{\left(3+\frac{3{x}^{2}}{3}+\frac{\left(3\cdot 2{x}^{4}\right)}{15}+\frac{\left(3\cdot 17{x}^{6}\right)}{315}+\frac{\left(3\cdot 62{x}^{8}\right)}{2835}+...\right)}$
Apply limit, x tends to 0
$=3/3$
$⇒1$
Result: 1