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If $\lim _{x \rightarrow 2} \frac{\tan \left(x-2\left\{x^2+k+2 x-2 k\right\}\right.}{x^2-4 x+4}=5$, then $\mathrm{k}$ is equal to:
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Verified Answer
The correct answer is:
3
3
$$
\begin{aligned}
\text { } & \lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+(k-2) x-2 k\right\}}{x^2-4 x+4}=5 \\
\Rightarrow & \lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+k x-2 x-2 k\right\}}{(x-2)^2}=5 \\
\Rightarrow & \lim _{x \rightarrow 2} \frac{\tan (x-2)\{x(x-2)+k(x-2)\}}{(x-2) \times(x-2)}=5 \\
\Rightarrow \lim _{x \rightarrow 2}\left(\frac{\tan (x-2)}{(x-2)}\right) \times \lim _{x \rightarrow 2}\left(\frac{(k+x)(x-2)}{(x-2)}\right)=5 \\
\Rightarrow & 1 \times \lim _{x \rightarrow 2}(k+x)=5 \\
&\left\{\because \lim _{h \rightarrow 0} \frac{\tan h}{h}=1\right\} \\
& \text { or } k+2=5 \\
\Rightarrow k=3
\end{aligned}
$$
\begin{aligned}
\text { } & \lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+(k-2) x-2 k\right\}}{x^2-4 x+4}=5 \\
\Rightarrow & \lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+k x-2 x-2 k\right\}}{(x-2)^2}=5 \\
\Rightarrow & \lim _{x \rightarrow 2} \frac{\tan (x-2)\{x(x-2)+k(x-2)\}}{(x-2) \times(x-2)}=5 \\
\Rightarrow \lim _{x \rightarrow 2}\left(\frac{\tan (x-2)}{(x-2)}\right) \times \lim _{x \rightarrow 2}\left(\frac{(k+x)(x-2)}{(x-2)}\right)=5 \\
\Rightarrow & 1 \times \lim _{x \rightarrow 2}(k+x)=5 \\
&\left\{\because \lim _{h \rightarrow 0} \frac{\tan h}{h}=1\right\} \\
& \text { or } k+2=5 \\
\Rightarrow k=3
\end{aligned}
$$
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