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Question Number 103366 by Study last updated on 14/Jul/20

	Answered by mathmax by abdo last updated on 14/Jul/20
	$let f(x) =((x^(1/4) −5^(1/{) )/((lnx)^(1/8) −(ln5)^(1/8) )) let find lim_(x→5) f(x) we do the changement x =5−t ⇒f(x) =g(t) =(((5−t)^(1/4) −5^(1/4) )/((ln(5−t))^(1/8) −(ln5)^(1/8) )) (t→0) =((5^(1/4) {(1−(t/5))^(1/4) −1})/((ln(5)+ln(1−(t/5)))^(1/8) −(ln5)^(1/8) )) ⇒g(t) ∼((5^(1/4) {1−(t/(20))−1})/((ln(5)−(t/5))^(1/8) −(ln5)^(1/8) )) =((5^(1/4) {−(t/(20))})/((ln5)^(1/8) {(1−(t/(5ln5)))^(1/8) −1})) ∼((5^(1/4) {−(t/(20))})/((ln5)^(1/8) {1−(t/(40ln5))−1})) →(5^(1/4) /((ln5)^(1/8) ))×((40ln5)/(20)) =((2 ln5 ×5^(1/4) )/((ln5)^(1/8) )) =2.5^(1/4) (ln5)^(7/8) ⇒lim_(x→5) f(x)= 2(^4 (√5))(ln5)^(7/8)$
	$let f (x) = \frac{x^{\frac{1}{4}} - 5^{\frac{1}{{}}}{{(lnx)}^{\frac{1}{8}} - {(\ln 5)}^{\frac{1}{8}}} let find \lim_{x \to 5} f (x) we do the changement$ $x = 5 - t \Rightarrow f (x) = g (t) = \frac{{(5 - t)}^{\frac{1}{4}} - 5^{\frac{1}{4}}}{{(\ln (5 - t))}^{\frac{1}{8}} - {(\ln 5)}^{\frac{1}{8}}} (t \to 0)$ $= \frac{5^{\frac{1}{4}} {{(1 - \frac{t}{5})}^{\frac{1}{4}} - 1}}{{(\ln (5) + \ln (1 - \frac{t}{5}))}^{\frac{1}{8}} - {(\ln 5)}^{\frac{1}{8}}}$ $\Rightarrow g (t) \sim \frac{5^{\frac{1}{4}} {1 - \frac{t}{20} - 1}}{{(\ln (5) - \frac{t}{5})}^{\frac{1}{8}} - {(\ln 5)}^{\frac{1}{8}}} = \frac{5^{\frac{1}{4}} {- \frac{t}{20}}}{{(\ln 5)}^{\frac{1}{8}} {{(1 - \frac{t}{5 \ln 5})}^{\frac{1}{8}} - 1}}$ $\sim \frac{5^{\frac{1}{4}} {- \frac{t}{20}}}{{(\ln 5)}^{\frac{1}{8}} {1 - \frac{t}{40 \ln 5} - 1}} \to \frac{5^{\frac{1}{4}}}{{(\ln 5)}^{\frac{1}{8}}} \times \frac{40 \ln 5}{20} = \frac{2 \ln 5 \times 5^{\frac{1}{4}}}{{(\ln 5)}^{\frac{1}{8}}}$ $= 2 . 5^{\frac{1}{4}} {(\ln 5)}^{\frac{7}{8}} \Rightarrow \lim_{x \to 5} f (x) = 2 (^{4} \sqrt{5}) {(\ln 5)}^{\frac{7}{8}}$
	Answered by bemath last updated on 14/Jul/20

	$(1) \lim_{x \to 5} \frac{\sqrt[4]{x} - \sqrt[4]{5}}{\sqrt[8]{\ln x} - \sqrt[8]{\ln 5}} =$ $\lim_{x \to 5} \frac{\frac{1}{4 x^{3 / 4}}}{\frac{1}{8 x {(\ln x)}^{7 / 8}}} = \lim_{x \to 5} \frac{8 x {(\ln x)}^{7 / 8}}{4 x^{3 / 4}}$ $= \frac{10 . {(\ln 5)}^{7 / 8}}{5^{3 / 4}} ◼$
	Answered by bemath last updated on 14/Jul/20

	$(2) \lim_{x \to 3} \frac{\sqrt[6]{e^{x}} - \sqrt[6]{e^{3}}}{\sqrt[3]{x} - \sqrt[3]{3}} =$ $\lim_{x \to 3} \frac{\frac{{(e^{x})}^{- 5 / 6} . e^{x}}{6}}{\frac{1}{3 \sqrt[3]{x^{2}}}} =$ $\lim_{x \to 3} \frac{3 \sqrt[3]{x^{2}}}{6} \times e^{\frac{x}{6}} = \frac{\sqrt[3]{9} . e^{1 / 2}}{2}$ $= \frac{\sqrt{e} . \sqrt[3]{9}}{2} ◼$
	Answered by mathmax by abdo last updated on 15/Jul/20
	$let g(x) =((e^(x/6) −e^(1/2) )/(x^(1/3) −3^(1/3) )) we do the cha7gement t =x−3 (so t→0) ⇒ g(x) =((e^((t+3)/6) −e^(1/2) )/((t+3)^(1/3) −3^(1/3) )) =((e^(1/2) {e^(t/6) −1})/(3^(1/3) { (1+(t/3))^(1/3) −1})) =ϕ(t) ⇒ ϕ(t) ∼(e^(1/2) /3^(1/3) ){ ((1+(t/6)−1)/(1+(t/9)−1))} →(e^(1/2) /3^(1/3) )×(9/6) =(3/2)×((√e)/((^3 (√3))))(t→0) ⇒ lim _(x→3) g(x) =((3(√e))/(2(^3 (√3))))$
	$let g (x) = \frac{e^{\frac{x}{6}} - e^{\frac{1}{2}}}{x^{\frac{1}{3}} - 3^{\frac{1}{3}}} we do the cha 7 gement t = x - 3 (so t \to 0) \Rightarrow$ $g (x) = \frac{e^{\frac{t + 3}{6}} - e^{\frac{1}{2}}}{{(t + 3)}^{\frac{1}{3}} - 3^{\frac{1}{3}}} = \frac{e^{\frac{1}{2}} {e^{\frac{t}{6}} - 1}}{3^{\frac{1}{3}} {{(1 + \frac{t}{3})}^{\frac{1}{3}} - 1}} = φ (t) \Rightarrow$ $φ (t) \sim \frac{e^{\frac{1}{2}}}{3^{\frac{1}{3}}} {\frac{1 + \frac{t}{6} - 1}{1 + \frac{t}{9} - 1}} \to \frac{e^{\frac{1}{2}}}{3^{\frac{1}{3}}} \times \frac{9}{6} = \frac{3}{2} \times \frac{\sqrt{e}}{(^{3} \sqrt{3})} (t \to 0) \Rightarrow$ $\lim_{x \to 3} g (x) = \frac{3 \sqrt{e}}{2 (^{3} \sqrt{3})}$
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