Question

Evaluate

$\displaystyle \lim_{x \to 0} \: \: \frac{tan \: x \: + \: 4tan2x \: - \: 3tan3x }{ {x}^{2}tan \: x }$
​

Answer 1

Solution

$\displaystyle \lim_{x \to 0} \: \: \frac{tan \: x \: + \: 4tan2x \: - \: 3tan3x }{ {x}^{2}tan \: x } \\ \\ = \displaystyle \lim_{x \to 0} \: \: \frac{tan \: x \: + \: 4( \dfrac{2tan \: x}{1 - {tan}^{2 } \: x } )\: - \: 3( \dfrac{3tan \: x \: - \: {tan}^{3} \: x}{1 - 3 {tan}^{2} x}) }{ {x}^{2}tan \: x } \\ \\ = \displaystyle \lim_{x \to 0} \dfrac{1 + \dfrac{8}{1 - {tan}^{2} x } - \dfrac{9 - 3 {tan}^{2}x }{1 - 3{tan}^{2}x} }{ {x}^{2} } \\ \\ = \displaystyle \lim_{x \to 0} \frac{(1 - {tan}^{2}x)(1 - 3 {tan}^{2}x) + 8(1 - 3{tan}^{2}x) - (9 - 3 {tan}^{2}x)(1 - {tan}^{2}x)}{ {x}^{2} (1 -{tan}^{2}x)(1 - 3 {tan}^{2}x) } \\ \\ = \displaystyle \lim_{x \to 0} \frac{1 - 4 {tan}^{2}x + 3{tan}^{4}x + 8 - 24 {tan}^{2}x - 9 + 12{tan}^{2}x - 3 {tan}^{4}x }{ {x}^{2} (1 -{tan}^{2}x)(1 - 3 {tan}^{2}x)} \\ \\ = \displaystyle \lim_{x \to 0} \: \frac{ - 16{tan}^{2}x}{ {x}^{2} (1 -{tan}^{2}x)(1 - 3 {tan}^{2}x)} \\ \\ = - 16 \: \displaystyle \lim_{x \to 0} \: ( \frac{tan \: x}{x}) {}^{2} \times \frac{1}{ (1 -{tan}^{2}x)(1 - 3 {tan}^{2}x)} \\ \\ = - 16 \times {1}^{2} \times \frac{1}{(1 - 0)(1 - 0)} \\ \\ = - 16$

Which is the required Answer.

Answer 2

See the step by step Explanation of this question

$\displaystyle \lim_{x \to 0} \: \: \frac{tan \: x \: + \: 4tan2x \: - \: 3tan3x }{ {x}^{2}tan \: x } \\ \\ = \displaystyle \lim_{x \to 0} \: \: \frac{tan \: x \: + \: 4( \dfrac{2tan \: x}{1 - {tan}^{2 } \: x } )\: - \: 3( \dfrac{3tan \: x \: - \: {tan}^{3} \: x}{1 - 3 {tan}^{2} x}) }{ {x}^{2}tan \: x } \\ \\ = \displaystyle \lim_{x \to 0} \dfrac{1 + \dfrac{8}{1 - {tan}^{2} x } - \dfrac{9 - 3 {tan}^{2}x }{1 - 3{tan}^{2}x} }{ {x}^{2} } \\ \\ = \displaystyle \lim_{x \to 0} \frac{(1 - {tan}^{2}x)(1 - 3 {tan}^{2}x) + 8(1 - 3{tan}^{2}x) - (9 - 3 {tan}^{2}x)(1 - {tan}^{2}x)}{ {x}^{2} (1 -{tan}^{2}x)(1 - 3 {tan}^{2}x) } \\ \\ = \displaystyle \lim_{x \to 0} \frac{1 - 4 {tan}^{2}x + 3{tan}^{4}x + 8 - 24 {tan}^{2}x - 9 + 12{tan}^{2}x - 3 {tan}^{4}x }{ {x}^{2} (1 -{tan}^{2}x)(1 - 3 {tan}^{2}x)} \\ \\ = \displaystyle \lim_{x \to 0} \: \frac{ - 16{tan}^{2}x}{ {x}^{2} (1 -{tan}^{2}x)(1 - 3 {tan}^{2}x)} \\ \\ = - 16 \: \displaystyle \lim_{x \to 0} \: ( \frac{tan \: x}{x}) {}^{2} \times \frac{1}{ (1 -{tan}^{2}x)(1 - 3 {tan}^{2}x)} \\ \\ = - 16 \times {1}^{2} \times \frac{1}{(1 - 0)(1 - 0)} \\ \\ = - 16$