lim ⁡ x → ∞ 5 x + 4 − 2 x − 1 = . . . \lim_{x\rightarrow\infty}\ \sqrt{5x+4}-\sqrt{2x-1}=\ ... x → ∞ lim 5 x + 4 <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 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"> = . . .

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">

5 \sqrt{5} 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">

lim ⁡ x → ∞ ( ( 2 x − 1 ) ( 3 x + 2 ) − ( 2 x − 5 ) ( 3 x + 5 ) ) = . . . \lim_{x\rightarrow\infty}\ \left(\sqrt{\left(2x-1\right)\left(3x+2\right)}-\sqrt{\left(2x-5\right)\left(3x+5\right)}\right)=\ ... x → ∞ lim ( ( 2 x − 1 ) ( 3 x + 2 ) <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"> − ( 2 x − 5 ) ( 3 x + 5 ) <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"> ) = . . .

1 2 6 \frac{1}{2}\sqrt{6} 2 1 6 <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">

lim ⁡ x → ∞ ( 2 x − 1 ) ( x + 2 ) − ( x 2 + 1 ) = . . . \lim_{x\rightarrow\infty}\ \sqrt{\left(2x-1\right)\left(x+2\right)}-\left(x\sqrt{2}+1\right)=\ ... x → ∞ lim ( 2 x − 1 ) ( x + 2 ) <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 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 ) = . . .

1 4 ( 3 2 − 4 ) \frac{1}{4}\left(3\sqrt{2}-4\right) 4 1 ( 3 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"> − 4 )

− 1 2 ( 3 2 − 4 ) -\frac{1}{2}\left(3\sqrt{2}-4\right) − 2 1 ( 3 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"> − 4 )

− 1 4 ( 3 2 − 4 ) -\frac{1}{4}\left(3\sqrt{2}-4\right) − 4 1 ( 3 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"> − 4 )

1 2 ( 3 2 − 4 ) \frac{1}{2}\left(3\sqrt{2}-4\right) 2 1 ( 3 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"> − 4 )

Latihan Soal "Limit Tak Hingga"

Quiz

•

Mathematics

•

11th Grade

•

Practice Problem

•

Easy

Ibu Sulistyana

Used 51+ times

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Latihan Soal "Limit Tak Hingga"

15 questions

$\lim_{x\rightarrow\infty}\ 5x\ =...$

$\lim_{x\rightarrow\infty}\ \left(-2x\right)=\ ...$

$\lim_{x\rightarrow\infty}\ \left(3x-1\right)=\ ...$

$\lim_{x\rightarrow\infty}\ \left(x^3-7x^2\right)\ =\ ...$

$\lim_{x\rightarrow\infty}\ \frac{2x^3+3x^2-5x+4}{2x^4-4x^2+9}=\ ...$

$\lim_{x\rightarrow\infty}\ \frac{2x^3+3x^2+7}{x^2+3x+4}=\ ...$

$\lim_{x\rightarrow\infty}\ \left(3-x+\frac{x^2-2x}{x+5}\right)\ =\ ...$

$\lim_{x\rightarrow\infty}\ \sqrt{5x+4}-\sqrt{2x-1}=\ ...$

$\lim_{x\rightarrow\infty}\ \sqrt{x+1}-\sqrt{x+2}=\ ...$

$\lim_{x\rightarrow\infty}\left(\sqrt{x-\sqrt{x}}-\sqrt{x+\sqrt{x}}\right)=\ ...$

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Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

Latihan Soal "Limit Tak Hingga"

15 questions

lim⁡x→∞ 5x =...\lim_{x\rightarrow\infty}\ 5x\ =...x→∞lim​ 5x =...

lim⁡x→∞ (−2x)= ...\lim_{x\rightarrow\infty}\ \left(-2x\right)=\ ...x→∞lim​ (−2x)= ...

lim⁡x→∞ (3x−1)= ...\lim_{x\rightarrow\infty}\ \left(3x-1\right)=\ ...x→∞lim​ (3x−1)= ...

lim⁡x→∞ (x3−7x2) = ...\lim_{x\rightarrow\infty}\ \left(x^3-7x^2\right)\ =\ ...x→∞lim​ (x3−7x2) = ...

lim⁡x→∞ 2x3+3x2−5x+42x4−4x2+9= ...\lim_{x\rightarrow\infty}\ \frac{2x^3+3x^2-5x+4}{2x^4-4x^2+9}=\ ...x→∞lim​ 2x4−4x2+92x3+3x2−5x+4​= ...

lim⁡x→∞ 2x3+3x2+7x2+3x+4= ...\lim_{x\rightarrow\infty}\ \frac{2x^3+3x^2+7}{x^2+3x+4}=\ ...x→∞lim​ x2+3x+42x3+3x2+7​= ...

lim⁡x→∞ (3−x+x2−2xx+5) = ...\lim_{x\rightarrow\infty}\ \left(3-x+\frac{x^2-2x}{x+5}\right)\ =\ ...x→∞lim​ (3−x+x+5x2−2x​) = ...

lim⁡x→∞ 5x+4−2x−1= ...\lim_{x\rightarrow\infty}\ \sqrt{5x+4}-\sqrt{2x-1}=\ ...x→∞lim​ 5x+4​−2x−1​= ...

lim⁡x→∞ x+1−x+2= ...\lim_{x\rightarrow\infty}\ \sqrt{x+1}-\sqrt{x+2}=\ ...x→∞lim​ x+1​−x+2​= ...

lim⁡x→∞(x−x−x+x)= ...\lim_{x\rightarrow\infty}\left(\sqrt{x-\sqrt{x}}-\sqrt{x+\sqrt{x}}\right)=\ ...x→∞lim​(x−x​​−x+x​​)= ...

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Popular Resources on Wayground

Discover more resources for Mathematics

$\lim_{x\rightarrow\infty}\ 5x\ =...$

$\lim_{x\rightarrow\infty}\ \left(-2x\right)=\ ...$

$\lim_{x\rightarrow\infty}\ \left(3x-1\right)=\ ...$

$\lim_{x\rightarrow\infty}\ \left(x^3-7x^2\right)\ =\ ...$

$\lim_{x\rightarrow\infty}\ \frac{2x^3+3x^2-5x+4}{2x^4-4x^2+9}=\ ...$

$\lim_{x\rightarrow\infty}\ \frac{2x^3+3x^2+7}{x^2+3x+4}=\ ...$

$\lim_{x\rightarrow\infty}\ \left(3-x+\frac{x^2-2x}{x+5}\right)\ =\ ...$

$\lim_{x\rightarrow\infty}\ \sqrt{5x+4}-\sqrt{2x-1}=\ ...$

$\lim_{x\rightarrow\infty}\ \sqrt{x+1}-\sqrt{x+2}=\ ...$

$\lim_{x\rightarrow\infty}\left(\sqrt{x-\sqrt{x}}-\sqrt{x+\sqrt{x}}\right)=\ ...$