lim _{x rightarrow 0} frac{tan (x)+4 tan (2 x | vedclass

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(D) We are given the limit: $\lim _{x ightarrow 0} \frac{	an x+4 	an 2 x-3 	an 3 x}{x^2 	an x}$.Using the expansion $	an x = x + \frac{x^3}{3}

Vedclass · Accepted Answer

$-16$

Vedclass · Answer

(D) We are given the limit: $\lim _{x ightarrow 0} \frac{	an x+4 	an 2 x-3 	an 3 x}{x^2 	an x}$.Using the expansion $	an x = x + \frac{x^3}{3} + O(x^5)$,we have:$	an x = x + \frac{x^3}{3} + O(x^5)$$	an 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$$	an 3x = 3x + \frac{(3x)^3}{3} + O(x^5) = 3x + 9x^3 + O(x^5)$Substituting these into the numerator:Numerator $= (x + \frac{x^3}{3}) + 4(2x + \frac{8x^3}{3}) - 3(3x + 9x^3) + O(x^5)$$= x + \frac{x^3}{3} + 8x + \frac{32x^3}{3} - 9x - 27x^3 + O(x^5)$$= (1+8-9)x + (\frac{1}{3} + \frac{32}{3} - 27)x^3 + O(x^5)$$= 0x + (\frac{33}{3} - 27)x^3 + O(x^5) = (11 - 27)x^3 = -16x^3 + O(x^5)$The denominator is $x^2 	an x \approx x^2(x) = x^3$.Thus,the limit is $\lim _{x ightarrow 0} \frac{-16x^3}{x^3} = -16$.

$\lim _{x \rightarrow 0} \frac{\tan (x)+4 \tan (2 x)-3 \tan (3 x)}{x^2 \tan (x)}$ is equal to

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