[tex]\sf\Large\lim\limits_{x\to0}\frac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}=...[/tex] A. 0 B. [tex]\sf\fra.... Question from @Corvusion

KLF

Jawab:
$\displaystyle D.\:\:\:\bf\frac{3}{2}$

Penjelasan dengan langkah-langkah:
Tentukan hasil dari
$\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}$
--> Caranya pake L'Hôpital
$\displaystyle\lim _{x\to 0}\frac{f'(x)}{g'(x)}$
dengan
$\displaystyle f(x)=\sqrt{1+x}-1$
dan
$\displaystyle g(x)=\sqrt[3]{1+x}-1$
--> maka
$\displaystyle\lim _{x\to 0}\frac{(\sqrt{1+x}-1)'}{(\sqrt[3]{1+x}-1)'}=\\\\\lim _{x\to 0}\frac{\sqrt{1+x}'-1'}{\sqrt[3]{1+x}'-1'}$
--> 1' = 0, krn c' = 0, maka . . .
$\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x}'}{\sqrt[3]{1+x}'}=\\\\\lim _{x\to 0}\frac{(1+x)^{\frac{1}{2}}'}{(1+x)^{\frac{1}{3}}'}$
--> Turunan.. (axⁿ)' = (na)x⁽ⁿ⁻¹⁾
$\displaystyle\lim _{x\to 0}\frac{(1+x)^{\frac{1}{2}}'}{(1+x)^{\frac{1}{3}}'}=\\\\\lim _{x\to 0}\frac{\frac{1}{2}(1+x)^{\frac{1}{2}-1}}{\frac{1}{3}(1+x)^{\frac{1}{3}-1}}=\\\\\lim _{x\to 0}\frac{\frac{1}{2}(1+x)^{\frac{1}{2}-\frac{2}{2}}}{\frac{1}{3}(1+x)^{\frac{1}{3}-\frac{3}{3}}}=\\\\\lim _{x\to 0}\frac{\frac{1}{2}(1+x)^{-\frac{1}{2}}}{\frac{1}{3}(1+x)^{-\frac{2}{3}}}=$
--> Karena
$\displaystyle x^{-n} = \frac{1}{x^n}\therefore\\\\\lim _{x\to 0}\left(\frac{1}{2(1+x)^{\frac{1}{2}}}\div\frac{1}{3(1+x)^{\frac{2}{3}}}\right)=\\\\\lim _{x\to 0}\left(\frac{1}{2\sqrt{1+x}}\div\frac{1}{3(1+x)^{\frac{2}{3}}}\right)$
--> x = 0, maka
$\displaystyle\frac{1}{2\sqrt{1+0}}\div\frac{1}{3(1+0)^{\frac{2}{3}}}=\\\\\frac{1}{2\sqrt{\not1}}\div\frac{1}{3(\not1)^{\not\frac{2}{3}}}=\\\\\frac{1}{2}\div\frac{1}{3}=\\\\\bf\frac{3}{2}$

[[ KLF ]]

4 votes Thanks 6

KLF Gak rapih. Aku fix dulu ya

lonerman latex nya ada yg blm :l

AlexiXiefer

»Limit Fungsi

$\sf \lim \limits_{x \to0} \frac{ \sqrt{1 + x} - 1}{ \sqrt[3]{1 + x} - 1} & = & \sf \lim \limits_{x \to0} \frac{(1 + x) {}^{ \frac{1}{2} } - 1 }{(1 + x) {}^{ \frac{1}{3} } - 1 } \\ & = & \sf\lim\limits_{x \to0} \frac{ \frac{1}{2}(1 + x) {}^{ \frac{1}{2} - 1 } }{ \frac{1}{3}(1 + x) {}^{ \frac{1}{3} - 1 } } \\ & = & \sf \lim \limits_{x \to0} \frac{ \frac{1}{2} (1 + x) {}^{ - \frac{1}{2} } }{ \frac{1}{3} (1 + x) {}^{ - \frac{2}{3} } } \\ & = & \sf \frac{ \frac{1}{2} (1 + 0) {}^{ - \frac{1}{2} } }{ \frac{1}{3} (1 + 0) {}^{ - \frac{2}{3} } } \\ & = & \sf \frac{ \frac{1}{2} }{ \frac{1}{3} } \\ & = & \bf{ \blue{ \boxed{ \frac{3}{2}}}}$

5 votes Thanks 24

Procyonion kok mirip jwabanku ya

AlexiXiefer ?.

Corvusion banh Eric kah?

Rixxel17 banh ....//cara pakai latex nya beda jdi beda lah

Procyonion cara latexnya beda² biar gk ketahuan kan?

Rixxel17 gk baik loh kk menuduh tanpa ad bukti yg kuat :v

»Limit Fungsi

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