Question

Evaluate the limit

$lim \: x \to \: \frac{\pi}{4} \: \: \frac{ {tan}^{3}x - tanx }{cos(x + \frac{\pi}{4}) }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{x \to \dfrac{\pi}{4}}\rm \: \: \frac{ {tan}^{3}x - tanx }{cos\bigg(x + \dfrac{\pi}{4} \bigg) }$

If we substitute directly the value of x, we get

$\rm \: = \: \dfrac{ {tan}^{3}\dfrac{\pi}{4} - tan\dfrac{\pi}{4} }{cos\bigg(\dfrac{\pi}{4} + \dfrac{\pi}{4} \bigg) }$

$\rm \: = \: \dfrac{ {1}^{3} - 1}{cos\dfrac{\pi}{2}}$

$\rm \: = \: \dfrac{1 - 1}{0}$

$\rm \: = \: \dfrac{0}{0}$

which is indeterminant form.

Consider, again

$\rm \: \displaystyle\lim_{x \to \dfrac{\pi}{4}}\rm \: \: \frac{ {tan}^{3}x - tanx }{cos\bigg(x + \dfrac{\pi}{4} \bigg) }$

$\rm \: = \: \displaystyle\lim_{x \to \dfrac{\pi}{4}}\rm \: \: \frac{tanx( {tan}^{2}x - 1)}{cos\bigg(x + \dfrac{\pi}{4} \bigg) }$

$\rm \: = \: \displaystyle\lim_{x \to \dfrac{\pi}{4}}\rm \: \: \frac{tanx( tanx + 1)(tanx - 1)}{cos\bigg(x + \dfrac{\pi}{4} \bigg) }$

$\rm \: = \: \displaystyle\lim_{x \to \dfrac{\pi}{4}}\rm tanx(tanx + 1) \times \displaystyle\lim_{x \to \dfrac{\pi}{4}}\rm \dfrac{tanx + 1}{cos\bigg(x + \dfrac{\pi}{4} \bigg) }$

$\rm \: = \: tan\dfrac{\pi}{4}\bigg(tan\dfrac{\pi}{4} + 1\bigg) \times \displaystyle\lim_{x \to \dfrac{\pi}{4}}\rm \frac{tanx + 1}{cos\bigg(x + \dfrac{\pi}{4} \bigg) }$

Now, we method of Substitution to solve this.

So, Substitute

$\rm \: x = \dfrac{\pi}{4} - h, \: \: as \: x \: \to \: \dfrac{\pi}{4} \: \: so \: h \: \to \: 0$

So, above expression can be rewritten as

$\rm \: = \: 1(1 + 1)\displaystyle\lim_{h \to 0}\rm \frac{tan\bigg(\dfrac{\pi}{4} - h \bigg) - 1}{cos\bigg(\dfrac{\pi}{4} + \dfrac{\pi}{4} - h \bigg) }$

$\rm \: = \: 2 \: \displaystyle\lim_{h \to 0}\rm \dfrac{\dfrac{tan\dfrac{\pi}{4} - tanh}{1 + tan\dfrac{\pi}{4}tanh} - 1}{sinh}$

$\rm \: = \: 2 \: \displaystyle\lim_{h \to 0}\rm \dfrac{\dfrac{1 - tanh}{1 + tanh} - 1}{sinh}$

$\rm \: = \: 2 \: \displaystyle\lim_{h \to 0}\rm \dfrac{\dfrac{1 - tanh - 1 - tanh}{1 + tanh}}{sinh}$

$\rm \: = \: 2 \: \displaystyle\lim_{h \to 0}\rm \dfrac{\dfrac{ - 2tanh}{1 + tanh}}{sinh}$

$\rm \: = \: - 4 \: \displaystyle\lim_{h \to 0}\rm \frac{1}{1 + tanh} \times \displaystyle\lim_{h \to 0}\rm \frac{tanh}{sinh}$

$\rm \: = \: - 4 \: \times \dfrac{1}{1 + tan0} \times \displaystyle\lim_{h \to 0}\rm \frac{sinh}{cosh} \times \frac{1}{sinh}$

$\rm \: = \: - 4 \: \times \dfrac{1}{1 + 0} \times \displaystyle\lim_{h \to 0}\rm \frac{1}{cosh}$

$\rm \: = \: - 4 \times 1 \times 1$

$\rm \: = \: - 4$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \: \displaystyle\lim_{x \to \dfrac{\pi}{4}}\rm \: \: \frac{ {tan}^{3}x - tanx }{cos\bigg(x + \dfrac{\pi}{4} \bigg) } = \: - \: 4 \: \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ \: \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1\: \: }}$

$\boxed{\tt{ \: \: \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1\: \: }}$

$\boxed{\tt{ \: \: \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1\: \: }}$

$\boxed{\tt{ \: \: \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1\: \: }}$

$\boxed{\tt{ \: \: \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga\: \: }}$