lim {((x^2 +5x+3)/(x^2 +x+2))}^x x→0 lim ((10^x −2^x −5^x +1)/(xtanx)) x→0


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Question Number 107828 by Rohit@Thakur last updated on 12/Aug/20
$lim {((x^2 +5x+3)/(x^2 +x+2))}^x x→0 lim ((10^x −2^x −5^x +1)/(xtanx)) x→0$
$l i m {\frac{x^{2} + 5 x + 3}{x^{2} + x + 2}}^{x}$ $x \to 0$ $l i m \frac{10^{x} - 2^{x} - 5^{x} + 1}{x t a n x}$ $x \to 0$
	Answered by Dwaipayan Shikari last updated on 12/Aug/20

	$\lim_{x \to 0} \frac{(5^{x} - 1) (2^{x} - 1)}{x s i n x} . c o s x$ $\lim_{x \to 0} \frac{(5^{x} - 1) (2^{x} - 1)}{x . x} c o s x (s i n x \to x)$ $\lim_{x \to 0} \frac{5^{x} - 1}{x} . \frac{2^{x} - 1}{x} = l o g 5 l o g 2$
	Answered by mathmax by abdo last updated on 13/Aug/20
	$f(x) ={((x^2 +5x +3)/(x^2 +x+2))}^x ⇒f(x) =e^(xln(((x^2 +5x+3)/(x^2 +x+2)))) ⇒f is continue at 0 and lim_(x→0) f(x) =e^0 =1$
	$f (x) = {\frac{x^{2} + 5 x + 3}{x^{2} + x + 2}}^{x} \Rightarrow f (x) = e^{xln (\frac{x^{2} + 5 x + 3}{x^{2} + x + 2})} \Rightarrow f is continue at 0$ $and \lim_{x \to 0} f (x) = e^{0} = 1$
	Answered by mathmax by abdo last updated on 13/Aug/20

	$g (x) = \frac{10^{x} - 2^{x} - 5^{x} + 1}{xtanx} \Rightarrow g (x) = \frac{e^{xln (10)} - e^{xln (2)} - e^{xln (5)} + 1}{xtanx}$ $let use hospital theorem \Rightarrow$ $\lim_{x \to 0} g (x) = \lim_{x \to 0} \frac{\ln (10) e^{xln 10} - \ln (2) e^{2 \ln 2} - \ln 5 e^{xln 5}}{tanx + x (1 + \tan^{2} x)}$ $= \lim_{x \to 0} \frac{\ln^{2} (10) e^{xln 10} - \ln^{2} (2) e^{2 \ln 2} - \ln^{2} 5 e^{xln 5}}{1 + \tan^{2} x + 1 + \tan^{2} x + x (. . . .)}$ $= \frac{\ln^{2} (10) - \ln^{2} (2) - \ln^{2} 5}{2} = \frac{{(\ln (2) + \ln 5)}^{2} - \ln^{2} 2 - \ln^{2} 5}{2}$ $= \frac{2 \ln (10)}{2} = \ln (10)$
	Commented by Dwaipayan Shikari last updated on 13/Aug/20

	Commented by Dwaipayan Shikari last updated on 13/Aug/20

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