Question

lim x =>0 4^x/2 + 4^-x/2 -2 / x^2
without using L hospital law
​

Answer 1

$\huge\boxed{\underline{\underline{\green{Solution}}}}$

$\displaystyle \lim_{x \to 0} \: \: \frac{ {4}^{ \frac{x}{2} } + {4}^{ - \frac{x}{2} } - 2 }{ {x}^{2} }$

$= \displaystyle \lim_{x \to 0} \: \: \frac{ { ({2}^{2}) }^{ \frac{x}{2} } + { ({2}^{2}) }^{ - \frac{x}{2} } - 2 }{ {x}^{2} }$

$= \displaystyle \lim_{x \to 0} \: \: \frac{ {2}^{ x } + {2}^{ - x } - 2 }{ {x}^{2} } \: \: \: ( \: \: \frac{0}{0} \: form \: )$

$= \displaystyle \lim_{x \to 0} \: \: \frac{ {2}^{ x }log2 - {2}^{ - x }log2 }{ 2{x} } \: \: \:$

$= \displaystyle \frac{log2}{2} \times \lim_{x \to 0} \: \: \frac{ {2}^{ x } - {2}^{ - x }}{ {x} } \: \: \: ( \: \: \frac{0}{0} \: form \: )$

$= \displaystyle \frac{ {(log2)} }{2}\lim_{x \to 0} \: \: \frac{ {2}^{ x }log2 + {2}^{ - x }log2 }{ 1} \: \: \:$

$= \displaystyle \frac{ {(log2)}^{2} }{4}\lim_{x \to 0} \: \: { ({2}^{ x } + {2}^{ - x }) } \: \: \:$

$= \displaystyle \frac{ {(log2)}^{2} }{4}(2 + 2)$

$= \displaystyle \frac{ {(log2)}^{2} }{4} \times 4$

$= \displaystyle { {(log2)}^{2} }$

$\displaystyle\textcolor{red}{Please \: Mark \: it \: Brainliest}</p><p></p><p>$