Question

lim x tends to pi/2 tan 3x / tan x

Answer 1

Final Answer :1/3

Steps:
1)
$\tan(3x) = \frac{3 \tan(x) - { (\tan(x) )}^{3} }{1 - 3 {( \tan(x) )}^{2} }$
2) Use above identity in given limits equation.

3) Solve limit.


For Calculation see Pic

Answer 2

The evaluated the expression lim x tends to $\frac{\pi}{2}$ ($\frac{tan3x}{tanx}$) is $\frac{1}{3}$ when the limits are applied.

Given that,

The expression is lim x tends to $\frac{\pi}{2}$ ($\frac{tan3x}{tanx}$)

We have to find the expansion.

We know that,

Take the expression

lim x tends to $\frac{\pi}{2}$ ($\frac{tan3x}{tanx}$)

Here we have tan3x the formula is $\frac{(3tanx-tan^{3} x)}{(1-3tan^{2} x)}$

Substitute in the limit

lim x tends to $\frac{\pi}{2}$ $\frac{(\frac{(3tanx-tan^{3} x)}{(1-3tan^{2} x)})}{tanx}$

Taking tanx common from numerator

lim x tends to $\frac{\pi}{2}$ $\frac{tanx(\frac{(3-tan^{2} x)}{(1-3tan^{2} x)})}{tanx}$

Canceling the numerator tanx and denominator tanx.

lim x tends to $\frac{\pi}{2}$ $\frac{(3-tan^{2} x)}{(1-3tan^{2} x)}$

Taking tan²x out we get

lim x tends to $\frac{\pi}{2}$ $\frac{tan^{2} x(\frac{3}{tan^{2} x} -1)}{tan^{2} x(\frac{1}{tan^{2} x} -3)}$

Canceling the numerator tan²x and denominator tan²x.

lim x tends to $\frac{\pi}{2}$ $\frac{(\frac{3}{tan^{2} x} -1)}{(\frac{1}{tan^{2} x} -3)}$

Applying limit

($\frac{3}{\infty}$-1)/($\frac{1}{\infty}$-3)

(0-1)/(0-3)

$\frac{1}{3}$

Therefore, The evaluated the expression lim x to $\frac{\pi}{2}$ ($\frac{tan3x}{tanx}$) is $\frac{1}{3}$  when the limits are applied.

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