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Soal dan Pembahasan limit fungsi aljabar

Pada kesempatan ini admin berbagi soal dan pembahasan limit fungsi aljabar. Soalnya akan diurutkan dari yang sederhana ke yang lebih kompleks. Sehingga sangat bagus bagi kalian yang baru belajar Bab limit atau membutuhkan sumber untuk belajar limit dari dasar sampai ke tingkat soal ujian nasional dan seleksi masuk perguruan tinggi nanti nya.
Download File Bab Limit Fungsi
Soal 1
$\displaystyle \lim_{x \to 9} 2x+7=2\times9+7=25$

Soal 2
$\displaystyle \lim_{x \to 5} (x^{2}+2x+1)=5^{2}+2\times5+1=36$

Soal 3
$\displaystyle \lim_{x \to 4} (x^{2}-10)^{3}=(4^{2}-10)^{3}=6^{3}=216$

Soal 4
$\displaystyle \lim_{x \to 3}(x+5)(2x-7)=(3+5)(2\times3-7)=-8$

Soal 5
$\displaystyle \lim_{x \to 4}\frac{x-4}{2\sqrt{x}-3}=\frac{0}{2\sqrt{4}-3}=0$
 
Soal 6
diketahui $\displaystyle \lim_{x \to 2} 2x^{2}-px+5=-1$ . Tentukan nilai p.
penyelesaian:
$\begin{align*}
2(2)^{2}-2p+5&=-1\\
8-2p+5&=-1\\
-2p&=-14\\
p&=7
\end{align*}$

Soal 7
$\displaystyle \lim_{x \to 4}\frac{x^{2}-2x-8}{x-4}$
penyelesaian:
$\displaystyle \lim_{x \to 4}\frac{x^{2}-2x-8}{x-4}=\frac{0}{0}$  (tak tentu)

$\begin{align*}\displaystyle \lim_{x \to 4}\frac{x^{2}-2x-8}{x-4}&=\displaystyle \lim_{x \to 4}\frac{(x-4)(x+2)}{x-4}\\
&=\displaystyle \lim_{x \to 4}x+2\\
&= \displaystyle \lim_{x \to 4}4+2\\
&=6
\end{align*}$

soal 8
$\displaystyle \lim_{x \to 1}\frac{2x^{2}-x-1}{3x^{2}-x-2}$
 penyelesaian:
$\displaystyle \lim_{x \to 1}\frac{2x^{2}-x-1}{3x^{2}-x-2}=\lim_{x \to 1}\frac{2\times1^{2}-1-1}{3\times1^{2}-1-2}=\frac{0}{0}$  (tak tentu)

$\begin{align*}
\displaystyle \lim_{x \to 1}\frac{2x^{2}-x-1}{3x^{2}-x-2}&=\lim_{x \to 1}\frac{(2x+1)(x-1)}{(x-1)(3x+2)}\\
&=\lim_{x \to 1}\frac{2x+1}{3x+2}\\
&=\frac{3}{5}
\end{align*}$ 

soal 9
$\displaystyle \lim_{x \to 5}\frac{x^{2}-9x+20}{x-5}$
penyelesaian:
$\displaystyle \lim_{x \to 5}\frac{x^{2}-9x+20}{x-5}=\frac{5^{2}-9\times5+20}{5-5}=\frac{0}{0}$  (tak tentu)

$\begin{align*}
 \displaystyle \lim_{x \to 5}\frac{x^{2}-9x+20}{x-5}&=\lim_{x \to 5}\frac{(x-5)(x-4)}{x-5}\\
&=\lim_{x \to 5}x-4\\
&=5-4=1
\end{align*}$

soal 10
 $\displaystyle \lim_{x \to 4}\frac{4-x}{2-\sqrt{x}}$
penyelesaian:
 $\displaystyle \lim_{x \to 4}\frac{4-x}{2-\sqrt{x}}=\frac{0}{0}$  (tak tentu)
$\begin{align*}
\displaystyle \lim_{x \to 4}\frac{4-x}{2-\sqrt{x}}&=\lim_{x \to 4}\frac{4-x}{2-\sqrt{x}}\times\frac{2+\sqrt{x}}{2+\sqrt{x}}\\
&= \lim_{x \to 4}\frac{(4-x)(2+\sqrt{x})}{4-x}\\
&= \lim_{x \to 4}2+\sqrt{x}\\
&=2+\sqrt{4}\\
&=4
\end{align*}$

soal 11
$\displaystyle \lim_{x \to 3}\frac{\sqrt{6x-2}-\sqrt{3x+7}}{x-3}$
penyelesaikan:
$\displaystyle \lim_{x \to 3}\frac{\sqrt{6x-2}-\sqrt{3x+7}}{x-3}=\frac{0}{0}$  (tak tentu)

$\begin{align*}
\displaystyle \lim_{x \to 3}\frac{\sqrt{6x-2}-\sqrt{3x+7}}{x-3}&=\displaystyle \lim_{x \to 3}\frac{\sqrt{6x-2}-\sqrt{3x+7}}{x-3}\times\frac{\sqrt{6x-2}+\sqrt{3x+7}}{\sqrt{6x-2}+\sqrt{3x+7}}\\
&=\displaystyle \lim_{x \to 3}\frac{(6x-2)-(3x+7)}{(x-3)(\sqrt{6x-2}+\sqrt{3x+7})}\\
&=\displaystyle \lim_{x \to 3}\frac{3x-9}{(x-3)(\sqrt{6x-2}+\sqrt{3x+7})}\\
&=\displaystyle \lim_{x \to 3}\frac{3(x-3)}{(x-3)(\sqrt{6x-2}+\sqrt{3x+7})}\\
&=\displaystyle \lim_{x \to 3}\frac{3}{\sqrt{6x-2}+\sqrt{3x+7}}\\
&=\frac{3}{\sqrt{16}+\sqrt{16}}\\
&=\frac{3}{8}
\end{align*}$

Soal 12 (UN 2018)
Diketahui $f(x)=\begin{cases}
 ax& \text{ if } x\leqslant 1 \\
 x+1& \text{ if } x> 1
\end{cases}$ . Agar $\displaystyle \lim_{x \to 1}f(x)$ mempunyai nilai, maka $a=...$
 penyelesaian:
syarat agar $f(x)$ mempunyai limit adalah nilai $f(x)$ ada  limit kiri=limit kanan. sehingga
$\begin{align*}\displaystyle \lim_{x \to 1}ax&=\displaystyle \lim_{x \to 1}x+1\\
 a\times1&=1+1\\
a&=2
\end{align*}$

Soal 13
$\displaystyle \lim_{x \to \infty}\sqrt{16x^{2}+10x-3}-4x+1$
penyelesaian:
 $\begin{align*}
\displaystyle \lim_{x \to \infty}\sqrt{16x^{2}+10x-3-2}-(4x-1)&=\displaystyle \lim_{x \to \infty}\sqrt{16x^{2}+10x-3-2}-\sqrt{(4x-1)^{2}}\\
&=\displaystyle \lim_{x \to \infty}\sqrt{16x^{2}+10x-3-2}-\sqrt{16x^{2}-8x+1}\\
&=\frac{10-(-8)}{2\sqrt{16}}\\
&=\frac{18}{8}\\
&=\frac{9}{4}
\end{align*}$

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