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Selamat datang di halaman contoh soal dan pembahasan "Limit".
Di halaman ini akan membahas tentang contoh soal dan pembahasan lengkap mengenai Limit.
Untuk lebih jelasnya mari kita lihat contoh soal dan pembahasan dibawah ini!
a. \(2\sqrt{2}\)
b. \(7\sqrt{2}\)
c. \(14\sqrt{2}\)
d. \(21\sqrt{2}\)
e. \(22\sqrt{2}\)
Pembahasan:
=\(\lim\limits_{x\to2}\frac{2x^{2}-x-6}{\sqrt{x}-\sqrt{2}}\cdot\frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\cdot +\sqrt{2}}\)
=\(\lim\limits_{x\to2}\frac{(2x+3)(x-2)\sqrt{x}+{2}}{(x-2)}\)
=\((2\cdot2+3)(\sqrt{2}+\sqrt{2})\)
=\(7\cdot2\sqrt{2}\)
=\(14\sqrt{2}\) (c.)
2. Hasil dari \(\lim\limits_{x\to1}\frac{1-x}{2-\sqrt{x+3}}\)=
a. 8
b. 4
c. 0
d. -4
e. -8
Pembahasan:
=\(\lim\limits_{x\to1}\frac{1-x}{2-\sqrt{x+3}}\cdot\frac{2+\sqrt{x+3}}{2+\sqrt{x+3}}\)
=\(\lim\limits_{x\to1}\frac{(1-x)\cdot(2+\sqrt{x+3})}{4-(x+3)}\)
=\(\lim\limits_{x\to1}\frac{(1-x)\cdot(2+\sqrt{x+3})}{(1-x)}\)
=\(2+\sqrt{1+3}\)
=\(2+\sqrt{4}=2+2=4\) (b.)
3. Hitunglah \(\lim\limits_{x\to1}\frac{\sqrt{x}-\sqrt{2x-1}}{x-1}\)=
a. \(\frac{1}{2}\)
b. \(\frac{3}{2}\)
c. \(-\frac{3}{2}\)
d. \(\frac{2}{3}\)
e. \(-\frac{1}{2}\)
Pembahasan:
Dalil L'Hospital
=\(\lim\limits_{x\to1}\frac{\frac{1}{2\sqrt{x}}-\frac{2}{2\sqrt{2x-1}}}{x-1}\)=
=\(\frac{\frac{1}{2}-\frac{1}{1}}{1}\)
=\(-\frac{1}{2}\) (e.)
4. Hasil dari \(\lim\limits_{x\to2}\frac{x-2}{\sqrt{3x-2}-\sqrt{2x}}\)=
a. 8
b. 4
c. 0
d. -4
e. -8
Pembahasan:
=\(\lim\limits_{x\to2}\frac{x-2}{\sqrt{3x-2}-\sqrt{2x}}\cdot\frac{\sqrt{3x-2}+\sqrt{2x}}{\sqrt{3x-2}+\sqrt{2x}}\)
=\(\lim\limits_{x\to2}\frac{(x-2)(\sqrt{3x-2}+\sqrt{2x})}{3x-2-2x}\)
=\(\lim\limits_{x\to2}\frac{(x-2)(\sqrt{3x-2}+\sqrt{2x})}{x-2}\)
=\(\lim\limits_{x\to2}(\sqrt{3x-2}+\sqrt{2x})\)
=\(\sqrt{3(2)-2}+\sqrt{2(2)}\)
=\(\sqrt{4}+\sqrt{4}=4\) (b.)
5. Hasil dari \(\lim\limits_{x\to3}\frac{(x^{2}-9)}{\sqrt{x^{2}+7}-4}\)=
a. 8
b. 4
c. 0
d. -4
e. -8
Pembahasan:
=\(\lim\limits_{x\to3}\frac{(x^{2}-9)}{\sqrt{x^{2}+7}-4}\cdot\frac{\sqrt{x^{2}+7}+4}{\sqrt{x^{2}+7}+4}\)
=\(\lim\limits_{x\to3}\frac{(x^{2}-9)(\sqrt{x^{2}+7}+4)}{(x^{2}+7)-16}\)
=\(\lim\limits_{x\to3}\frac{(x^{2}-9)(\sqrt{x^{2}+7}+4)}{(x^{2}-9}\)
=\(\lim\limits_{x\to3}(\sqrt{x^{2}+7}+4)=8\) (a.)
6. Hitunglah \(\lim\limits_{x\to0}\frac{\sqrt{x^{2}+x}-\sqrt{x}}{x\sqrt{x}}\)=
a. \(\frac{1}{2}\)
b. \(\frac{3}{2}\)
c. \(-\frac{3}{2}\)
d. \(\frac{2}{3}\)
e. \(-\frac{1}{2}\)
Pembahasan:
=\(\lim\limits_{x\to0}\frac{\sqrt{x^{2}+x}-\sqrt{x}}{x\sqrt{x}}\cdot\frac{\sqrt{x^{2}+x}+\sqrt{x}}{\sqrt{x^{2}+x}+\sqrt{x}}\)
=\(\lim\limits_{x\to0}\frac{x^{2}+x-x}{(x\sqrt{x})(\sqrt{x^{2}+x}+\sqrt{x})}\)
=\(\lim\limits_{x\to0}\frac{x^{2}}{(x\sqrt{x})(\sqrt{x}\sqrt{x+1}+\sqrt{x})}\)
=\(\lim\limits_{x\to0}\frac{x^{2}}{(x^{2})(\sqrt{x+1}+1)}\)
=\(\frac{1}{\sqrt{x+1}+1}\)
=\(\frac{1}{2}\) (a.)
7. Hitunglah \(\lim\limits_{x\to3}\frac{\sqrt{3x+7}-\sqrt{4x+4}}{2x-6}\)=
a. \(\frac{1}{16}\)
b. \(\frac{3}{2}\)
c. \(-\frac{1}{16}\)
d. \(\frac{2}{3}\)
e. \(-\frac{1}{2}\)
Pembahasan:
=\(\lim\limits_{x\to3}\frac{\sqrt{3x+7}-\sqrt{4x+4}}{2x-6}\cdot\frac{\sqrt{3x+7}+\sqrt{4x+4}}{{3x+7}+\sqrt{4x+4}}\)
=\(\lim\limits_{x\to3}\frac{(3x+7)-(4x-4)}{2(x-3)(\sqrt{3x+7}+\sqrt{4x+4})}\)
=\(\lim\limits_{x\to3}\frac{-x+3}{2(x-3)(\sqrt{3x+7}+\sqrt{4x+4})}\)
=\(\lim\limits_{x\to3}\frac{-1}{2(\sqrt{3x+7}+\sqrt{4x+4})}\)
=\(\frac{-1}{2(4+4)}\)
=\(-\frac{1}{16}\) (c.)
8. Tentukan nilai \(\lim\limits_{x\to3}\frac{x-\sqrt{2x+3}}{9-x^{2}}\)=
a. \(\frac{1}{16}\)
b. \(\frac{1}{9}\)
c. \(-\frac{1}{9}\)
d. \(\frac{2}{3}\)
e. \(-\frac{1}{2}\)
Pembahasan:
=\(\lim\limits_{x\to3}\frac{x-\sqrt{2x+3}}{9-x^{2}}\cdot\frac{x+\sqrt{2x+3}}{x+\sqrt{2x+3}}\)
=\(\lim\limits_{x\to3}\frac{x^{2}-2x-3}{(3-x)(3+3)(x+\sqrt{2x+3})}\)
=\(\lim\limits_{x\to3}\frac{(x-3)(x+1)}{-(3-x)(3+3)(x+\sqrt{2x+3}+x)}\)
=\(\frac{4}{-(6)(3+3)}\)
=\(-\frac{4}{36}\)
=\(-\frac{1}{9}\) (c.)
9. Hitunglah \(\lim\limits_{x\to3}\frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\)=
a. \(\frac{1}{14}\sqrt{7}\)
b. \(\frac{1}{9}\sqrt{7}\)
c. \(-\frac{1}{14}\sqrt{7}\)
d. \(\frac{2}{3}\sqrt{7}\)
e. \(-\frac{1}{2}\sqrt{7}\)
Pembahasan:
=\(\lim\limits_{x\to3}\frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\cdot\frac{\sqrt{x+4}+\sqrt{2x+1}}{\sqrt{x+4}+\sqrt{2x+1}}\)
=\(\lim\limits_{x\to3}\frac{x+4-(2x+1)}{(x-3)(\sqrt{x+4}+\sqrt{2x+1})}\)
=\(\lim\limits_{x\to3}\frac{-x+3}{(x-3)(\sqrt{x+4}+\sqrt{2x+1})}\)
=\(\lim\limits_{x\to3}\frac{-1}{\sqrt{x+4}+\sqrt{2x+1}}\)
=\(\frac{x+4-(2x+1)}{\sqrt{3+4}+\sqrt{2(3)+1}}\)
=\(\frac{-1}{\sqrt{7}+\sqrt{7}}\)
=\(\frac{-1}{2\sqrt{7}}\)
=\(-\frac{1}{14}\sqrt{7}\) (c)
10. Tentukan nilai \(\lim\limits_{x\to2}\frac{x^{2}-3x+2}{\sqrt{x-2}}\)=
a. 8
b. 4
c. 0
d. -4
e. -8
Pembahasan:
=\(\lim\limits_{x\to2}\frac{x^{2}-3x+2}{\sqrt{x-2}}\cdot\frac{\sqrt{x-2}}{\sqrt{x-2}}\)
=\(\lim\limits_{x\to2}\frac{(x^{2}-3x+2)(\sqrt{x-2})}{(\sqrt{x-2})^{2}}\)
=\(\lim\limits_{x\to2}\frac{(x-1)(x-2)(\sqrt{x-2})}{(x-2)}\)
=\(\lim\limits_{x\to2}(x-1)\sqrt{x-2}=0\) (c.)
