lim ⁡ x → ∞ x − 3 − 4 x + 1 2 x − x − 5 \lim_{x\rightarrow\infty}\ \frac{\sqrt{x-3}-\sqrt{4x+1}}{\sqrt{2x}-\sqrt{x-5}} x → ∞ lim 2 x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> − x − 5 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> x − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> − 4 x + 1 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> = ...

− 1 − 2 -1-\sqrt{2} − 1 − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 + 1 \sqrt{2}+1 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> + 1

2 \sqrt{2} 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 − 1 \sqrt{2}-1 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> − 1

lim ⁡ x → ∞ ( ( 2 x ) ( x − 1 ) − x ( x + 3 ) ) \lim_{x\rightarrow\infty}\ \left(\sqrt{\left(2x\right)\left(x-1\right)}-\sqrt{x\left(x+3\right)}\right) x → ∞ lim ( ( 2 x ) ( x − 1 ) <path d="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167, 158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067 c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84, 175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71 c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3, -252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26 s76,-59,76,-59s76,-60,76,-60z M1001 80H40000v40H1012z"> − x ( x + 3 ) <path d="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167, 158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067 c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84, 175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71 c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3, -252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26 s76,-59,76,-59s76,-60,76,-60z M1001 80H40000v40H1012z"> ) = ...

− 5 2 2 -\frac{5}{2}\sqrt{2} − 2 5 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 4 2 \frac{1}{4}\sqrt{2} 4 1 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 2 2 \frac{1}{2}\sqrt{2} 2 1 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

menentukan nilai limit tak hingga

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menentukan nilai limit tak hingga

Quiz

•

Mathematics

•

11th Grade

•

Practice Problem

•

Hard

•

CCSS

HSA.SSE.A.2, HSA.APR.A.1, HSA.SSE.B.3

Standards-aligned

Juwanah Ana

Used 214+ times

FREE Resource

46 questions

Show all answers

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Nilai dari $\lim_{x\rightarrow\infty}\left(\frac{1-5x}{2x-3}\right)\ =...$

$-\infty$

$-\frac{5}{2}$

$\frac{1}{2}$

$\frac{5}{3}$

$\infty$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Nilai dari $\lim_{x\rightarrow\infty}\left(\frac{1-5x}{2x-3}\right)\ =...$

$-\infty$

$-\frac{5}{2}$

$\frac{1}{2}$

$\frac{5}{3}$

$\infty$

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Nilai dari $\lim_{x\rightarrow\infty}\left(\frac{10x^2}{2x-1}-5x\right)\ =\ ....$

-10

$-\frac{5}{2}$

$\frac{5}{2}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Nilai dari $\lim_{x\rightarrow\infty}\left(\sqrt{x^2+7x+5}-\sqrt{x^2-2x+3}\right)$ adalah ....

$8\ \frac{1}{2}$

$7\ \frac{1}{2}$

$6\ \frac{1}{2}$

$5\ \frac{1}{2}$

$4\ \frac{1}{2}$

Nilai dari $\lim_{x\rightarrow\infty}\left(\sqrt{16x^2+10x-3}-4x+1\right)=$ ......

$-\frac{9}{4}$

$-\frac{1}{4}$

$\frac{1}{4}$

$\frac{5}{4}$

$\frac{9}{4}$

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Nilai dari $\lim_{x\rightarrow\infty}\left(2x-\sqrt{4x^2+x+3}\right)=....$

$-\frac{1}{2}$

$-\frac{1}{4}$

$0$

$\frac{1}{4}$

$\frac{1}{2}$

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menentukan nilai limit tak hingga

46 questions

Nilai dari lim⁡x→∞(1−5x2x−3) =...\lim_{x\rightarrow\infty}\left(\frac{1-5x}{2x-3}\right)\ =...x→∞lim​(2x−31−5x​) =...

Nilai dari lim⁡x→∞(1−5x2x−3) =...\lim_{x\rightarrow\infty}\left(\frac{1-5x}{2x-3}\right)\ =...x→∞lim​(2x−31−5x​) =...

Nilai dari lim⁡x→∞(10x22x−1−5x) = ....\lim_{x\rightarrow\infty}\left(\frac{10x^2}{2x-1}-5x\right)\ =\ ....x→∞lim​(2x−110x2​−5x) = ....

Nilai dari lim⁡x→∞(x2+7x+5−x2−2x+3)\lim_{x\rightarrow\infty}\left(\sqrt{x^2+7x+5}-\sqrt{x^2-2x+3}\right)x→∞lim​(x2+7x+5​−x2−2x+3​) adalah ....

Nilai dari lim⁡x→∞(16x2+10x−3−4x+1)=\lim_{x\rightarrow\infty}\left(\sqrt{16x^2+10x-3}-4x+1\right)=x→∞lim​(16x2+10x−3​−4x+1)= ......

Nilai dari lim⁡x→∞(2x−4x2+x+3)=....\lim_{x\rightarrow\infty}\left(2x-\sqrt{4x^2+x+3}\right)=....x→∞lim​(2x−4x2+x+3​)=....

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Nilai dari $\lim_{x\rightarrow\infty}\left(\frac{1-5x}{2x-3}\right)\ =...$

Nilai dari $\lim_{x\rightarrow\infty}\left(\frac{1-5x}{2x-3}\right)\ =...$

Nilai dari $\lim_{x\rightarrow\infty}\left(\frac{10x^2}{2x-1}-5x\right)\ =\ ....$

Nilai dari $\lim_{x\rightarrow\infty}\left(\sqrt{x^2+7x+5}-\sqrt{x^2-2x+3}\right)$ adalah ....

Nilai dari $\lim_{x\rightarrow\infty}\left(\sqrt{16x^2+10x-3}-4x+1\right)=$ ......

Nilai dari $\lim_{x\rightarrow\infty}\left(2x-\sqrt{4x^2+x+3}\right)=....$