lim_(x→0) ((2^(sec(x)) − 2^(cos(x)) )/x^2 )


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Question Number 63579 by Tawa1 last updated on 05/Jul/19

$\lim_{x \to 0} \frac{2^{\sec (x)} - 2^{\cos (x)}}{x^{2}}$
Commented by Prithwish sen last updated on 05/Jul/19

$∵ form \frac{0}{0} applying L^{'} Hopital$ $li \underset{x \to 0}{m} \frac{secx tanx 2^{secx} \ln 2 + sinx 2^{cosx} \ln 2}{2 x}$ $= \frac{1}{2} [li \underset{x \to 0}{m} (\frac{tanx}{x}) . secx 2^{secx} \ln 2 + li \underset{x \to 0}{m} (\frac{sinx}{x}) . 2^{cosx} \ln 2]$ $= 2 \ln 2 please check .$
Commented by mathmax by abdo last updated on 05/Jul/19
$let use hospital theorem let A(x)=((2^(1/(cosx)) −2^(cosx) )/x^2 ) u(x)=2^(1/(cosx)) −2^(cosx) and v(x)=x^2 we have lim_(x→0) u(x)=lim_(x→0) v(x)=0 u(x)=e^((ln(2))/(cosx)) −e^(ln(2)cosx) ⇒u^′ (x) =(((ln(2))/(cosx)))^′ e^((ln(2))/(cosx)) +ln(2)sinx e^(ln(2)cosx) =((ln(2)sinx)/(cos^2 x)) e^((ln(2))/(cosx)) + ln(2)sinx e^(ln(2) cosx) =ln(2)sinx{ (e^((ln(2))/(cosx)) /(cos^2 x)) +e^(ln(2)cosx) } ⇒ u^((2)) (x)=ln(2)cosx{ (e^((ln(2))/(cosx)) /(cos^2 x)) +e^(ln(2)cosx) } +ln(2)sinx {.......}^((1)) ⇒ lim_(x→0) u^((2)) (x) = ln(2){2 +2}=4ln(2) also we have v^′ (x)=2x and v^((2)) (x)=2 ⇒ lim_(x→0) v^((2)) (x)=2 ⇒lim_(x→0) A(x) =((4ln(2))/2) =2ln(2).$
$l e t u s e h o s p i t a l t h e o r e m l e t A (x) = \frac{2^{\frac{1}{c o s x}} - 2^{c o s x}}{x^{2}}$ $u (x) = 2^{\frac{1}{c o s x}} - 2^{c o s x} a n d v (x) = x^{2} w e h a v e {l i m}_{x \to 0} u (x) = {l i m}_{x \to 0} v (x) = 0$ $u (x) = e^{\frac{l n (2)}{c o s x}} - e^{l n (2) c o s x} \Rightarrow u^{^{'}} (x) = {(\frac{l n (2)}{c o s x})}^{^{'}} e^{\frac{l n (2)}{c o s x}} + l n (2) s i n x e^{l n (2) c o s x}$ $= \frac{l n (2) s i n x}{{c o s}^{2} x} e^{\frac{l n (2)}{c o s x}} + l n (2) s i n x e^{l n (2) c o s x} = l n (2) s i n x {\frac{e^{\frac{l n (2)}{c o s x}}}{{c o s}^{2} x} + e^{l n (2) c o s x}} \Rightarrow$ $u^{(2)} (x) = l n (2) c o s x {\frac{e^{\frac{l n (2)}{c o s x}}}{{c o s}^{2} x} + e^{l n (2) c o s x}} + l n (2) s i n x {. . . . . . .}^{(1)} \Rightarrow$ ${l i m}_{x \to 0} u^{(2)} (x) = l n (2) {2 + 2} = 4 l n (2) a l s o w e h a v e$ $v^{^{'}} (x) = 2 x a n d v^{(2)} (x) = 2 \Rightarrow {l i m}_{x \to 0} v^{(2)} (x) = 2 \Rightarrow {l i m}_{x \to 0} A (x) = \frac{4 l n (2)}{2} = 2 l n (2) .$
Commented by Tawa1 last updated on 05/Jul/19

$God bless you sir$
Commented by mathmax by abdo last updated on 05/Jul/19

$y o u a r e w e l c o m e .$
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