Question

solve this question guys,,,,,, see attachment ​

Answer 1

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

Given that

$\rm \: f(x) = \bigg[tan\bigg(\dfrac{\dfrac{\pi}{4} + ln(x) }{\dfrac{\pi}{4} - ln(x)} \bigg]^{ log_{x}(e) } \: is \: continuous \: at \: x = 1$

So, it means

$\rm \:\displaystyle\lim_{x \to \: 1}\rm \bigg[tan\bigg(\dfrac{\dfrac{\pi}{4} + ln(x) }{\dfrac{\pi}{4} - ln(x)} \bigg]^{ log_{x}(e) } = f(1)$

$\rm \:\displaystyle\lim_{x \to \: 1}\rm \bigg[tan\bigg(\dfrac{\dfrac{\pi}{4} + ln(x) }{\dfrac{\pi}{4} - ln(x)} \bigg]^{ log_{x}(e) } = {e}^{k} \cdots \cdots \: (1)$

Now, Consider

$\rm \:\displaystyle\lim_{x \to \: 1}\rm \bigg[tan\bigg(\dfrac{\dfrac{\pi}{4} + ln(x) }{\dfrac{\pi}{4} - ln(x)} \bigg]^{ log_{x}(e) }$

We know,

$\boxed{ \rm{ \:tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: }}$

So, using this, we get

$\rm \: = \: \displaystyle\lim_{x \to \: 1}\rm \bigg[\dfrac{tan\dfrac{\pi}{4} + tan(ln \: x)}{1 - tan\dfrac{\pi}{4} \times tan(ln \: x)} \bigg]^{ log_{x}(e) }$

$\rm \: = \: \displaystyle\lim_{x \to \: 1}\rm \bigg[\dfrac{1 + tan(ln \: x)}{1 - 1 \times tan(ln \: x)} \bigg]^{ log_{x}(e) }$

$\rm \: = \: \displaystyle\lim_{x \to \: 1}\rm \bigg[\dfrac{1 + tan(ln \: x)}{1 - tan(ln \: x)} \bigg]^{ log_{x}(e) }$

$\rm \: = \: \displaystyle\lim_{x \to \: 1}\rm \bigg[1 + \dfrac{1 + tan(ln \: x)}{1 - tan(ln \: x)} - 1 \bigg]^{ log_{x}(e) }$

$\rm \: = \: \displaystyle\lim_{x \to \: 1}\rm \bigg[1 + \dfrac{1 + tan(ln \: x) - 1 + tan(ln \: x)}{1 - tan(ln \: x)} \bigg]^{ log_{x}(e) }$

$\rm \: = \: \displaystyle\lim_{x \to \: 1}\rm \bigg[1 + \dfrac{2tan(ln \: x)}{1 - tan(ln \: x)} \bigg]^{ log_{x}(e) }$

$\rm \: = \: \displaystyle\lim_{x \to \: 1}\rm \bigg[1 + \dfrac{2tan(ln \: x)}{1 - tan(ln \: x)} \bigg]^{\dfrac{1 - tan(ln \: x)}{2tan(ln \: x)} \times \dfrac{2tan(ln \: x)}{1 - tan(ln \: x)} \times \dfrac{1}{ log_{e}(x) } }$

We know,

$\boxed{ \rm{ \:\displaystyle\lim_{x \to \: 0}\rm {\bigg(1 + x\bigg)}^{ \dfrac{1}{x} } = e\: }}$

So, using this, we get

$\rm \: = \: e^{\displaystyle\lim_{x \to \: 1}\rm \dfrac{2tan(ln \: x)}{1 - tan(ln \: x)} \times \dfrac{1}{ log_{e}(x) } }$

$\rm \: = \: e^{\displaystyle\lim_{x \to \: 1}\rm \dfrac{2tan(ln \: x)}{1 - tan(ln \: x)} \times \dfrac{1}{ ln \: (x) } }$

$\rm \: = \: e^{2 \: \displaystyle\lim_{x \to \: 1}\rm \dfrac{tan(ln \: x)}{ln \: x} \times \dfrac{1}{1 - tan( ln \: x) } }$

$\rm \: = \: {e}^{2 \times 1 \times 1}$

$\rm \: = \: {e}^{2}$

So, equation (1), can be rewritten as

$\rm \: {e}^{2} = {e}^{k}$

$\rm\implies \:k \: = \: 2$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

For continuity at x=1, we must have

$\[ \begin{aligned} \rm f(1) & \rm= \lim_{x \rightarrow 1} f(x) \\ \\ & \rm= \lim_{x \rightarrow 1}\left(\tan \left(\frac{\pi}{4}+\ell n x\right)\right)^{1 / \ell n x} \\ \\ & \rm=e^{ \displaystyle \rm\lim_{x \rightarrow 1}\frac{\left(\tan \left(\frac{\pi}{4}+\ln x\right)-1\right)}{\ell n x}} \\ \\ & \rm=e^{ \displaystyle \rm\lim_{x \rightarrow 1}\frac{1+\tan (\ln x)-(1-\tan (\ln x))}{(1-\tan (\ell n x)) \cdot \ell n x}} \\ \\ & \rm=e^{ \displaystyle \rm\lim_{x \rightarrow 1} \frac{2 \cdot \tan (\ln x)}{\ln x} \cdot \frac{1}{1-\tan (\ln x)}} \\ \\ & \boxed{\color{magenta} \rm=e^{2}} \end{aligned} \]$

So, equation (1), can be rewritten as

$\begin{gathered}\rm \: {e}^{2} = {e}^{k} \\ \end{gathered}$

$\begin{gathered} \boxed{\red{ \rm\implies \:k \: = \: 2 }}\\ \end{gathered}$