Let $g (x) = lim n \to \infty \frac{x^{n} f (x) + h (x) + 1}{2 x^{n} + 3 x + 3},$ $x^{1} 1$ and $g (1) = lim x \to 1 \frac{{sin}^{2} (π \cdot 2^{x})}{l n (sec (π \cdot 2^{x}))}$ be a continuous function at $x = 1$ , find the value of $4 g (1) + 2 f (1) - h (1)$ . Assume that $f (x)$ and $h (x)$ are continuous at $x = 1$ .

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Solution

$g (1) = {lim}_{x \to 1} \frac{{sin}^{2} (π . 2^{x})}{ln (sec (π . 2^{x}))} \frac{0}{0}$
form
$= {lim}_{x \to 0} \frac{sin (π . 2^{x}) cos (π . 2^{x}) (π . 2^{x} log 2)}{\frac{(π . 2^{x} log 2)}{sec (π . 2^{x})} \times sec (π . 2^{x}) tan (π . 2^{x})}$
$= {lim}_{x \to 0} 2 {cos}^{2} (π . 2^{x}) = 2 \times 1 = 2$
For $x > 1$
as $n \to \infty, x^{n} \to \infty$
$g (x) = {lim}_{n \to \infty} \frac{f (2) + \frac{ln x}{x^{n}} + \frac{1}{x^{n}}}{2 + \frac{3 x}{x^{n}} + \frac{3}{x^{n}}}$
$= \frac{f (x) + 0 + 0}{2 + 0 + 0} = \frac{f (x)}{2}$
For $x < 1$
$g (x) = {lim}_{n \to \infty} \frac{x^{n} f (x) + h (x) + 1}{2 x^{n} + 3 x + 3}$
$= \frac{0 + h (x) + 1}{0 + 3 x + 3} = \frac{h (x) + 1}{3 (x + 1)}$
${lim}_{x \to 1^{-}} g (x) = {lim}_{x \to 1^{-}} \frac{h (x) + 1}{3 (x + 1)} = \frac{h (1) + 1}{6}$
${lim}_{x \to 1^{+}} g (x) = {lim}_{x \to 1^{+}} \frac{f (x)}{2} = \frac{f (1)}{2}$
Given $g (x)$ is continuous at $x = 1$
$\Rightarrow g (1) = \frac{h (1) + 1}{6} = \frac{f (1)}{2}$
$\Rightarrow f (x) = 2 g (1), h (1) = 6 g (1) - 1$
$\Rightarrow g (1) + 2 f (1) - h (1)$
$= 4 g (1) + 2 (2 g (1)) - (6 g (1) - 1)$
$= 2 g (1) - 1 = 4 - 1 = 3$

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