Teorema Limit ~ Matematika SMA

Sebelumnya kita mengetahui bahwa, jika $f(a)$ terdefinisi, maka $\displaystyle\lim_{x\to a}f(x) = f(a)$. Dari sini, dapat dikembangkan menjadi beberapa teorema seperti berikut.

Teorema

Misalkan $n$ merupakan bilangan asli, $k$ konstanta, serta $f$ dan $g$ fungsi-fungsi yang mempunyai limit di $a$, maka

1. 
   $\displaystyle\lim_{x\to a}k = k$
   Contoh:$\displaystyle\lim_{x\to5}3 = 3$
2. 
   $\displaystyle\lim_{x\to a}kf(x) = k\lim_{x\to a}f(x)$
   Contoh:
   $\displaystyle\lim_{x\to 2}5(x+7)= 5\lim_{x\to 2}(x+7)=5(9)=45$
3. 
   $\displaystyle\lim_{x\to a}\left[f(x)\pm g(x)\right] = \lim_{x\to a}f(x)\pm\lim_{x\to a}g(x)$
   Contoh:
   $\displaystyle
   \begin{aligned}
   \lim_{x\to 2}\left[(x+7)+(2x-5)\right]&=\lim_{x\to 2}(x+7)+\lim_{x\to 2}(2x-5)\\
   \lim_{x\to 2} 3x+2 &=\lim_{x\to 2}(x+7)+\lim_{x\to 2}(2x-5)\\
   3(2)+2 &= (2+7) + (2(2)-5)\\
   8 &= 9 + (-1)\\
   8 &= 8\end{aligned}$
4. 
   $\displaystyle\lim_{x\to a}\left[f(x)\times g(x)\right] = \lim_{x\to a}f(x)\times\lim_{x\to a}g(x)$
   Contoh:
   $\displaystyle
   \begin{aligned}
   \lim_{x\to 2}\left[(x+7)(2x-5)\right]&=\lim_{x\to 2}(x+7)\times\lim_{x\to 2}(2x-5)\\
   \lim_{x\to 2} 2x^2+9x-35 &=\lim_{x\to 2}(x+7)\times\lim_{x\to 2}(2x-5)\\
   2(2)^2+9(2)-35 &= (2+7) \times (2(2)-5)\\
   8+18-35 &= 9 \times (-1)\\
   -9 &= -9\end{aligned}$
5. 
   $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ dengan syarat $\displaystyle\lim_{x\to a}g(x)\ne0$
   Contoh:
   $\displaystyle
   \begin{aligned}
   \lim_{x\to 2}\frac{x+7}{2x-5}&=\frac{\lim_{x\to 2}(x+7)}{\lim_{x\to 2}(2x-5)}\\
   \frac{2+7}{2(2)-5} &= \frac{2+7}{2(2)-5} \\
   -9 &= -9\end{aligned}$
6. 
   $\displaystyle\lim_{x\to a}\left[f(x)\right]^n= \left[\lim_{x\to a}f(x)\right]^n$
   Contoh:
   $\displaystyle
   \begin{aligned}
   \lim_{x\to 2}(x+7)^2&=\left(\lim_{x\to 2}(x+7)\right)^2\\
   \lim_{x\to 2}x^2+14x+49&=\left(\lim_{x\to 2}(x+7)\right)^2\\
   (2)^2+14(2)+49 &= (2+7)^2 \\
   81 &= 81\end{aligned}$
7. 
   $\displaystyle\lim_{x\to a}\sqrt[n]{f(x)}= \sqrt[n]{\lim_{x\to a}f(x)}$
   Contoh:
   $\displaystyle
   \begin{aligned}
   \lim_{x\to 2}\sqrt{x+7}&=\sqrt{\lim_{x\to 2}(x+7)}\\
   \sqrt{9} &= \sqrt{9} \\
   3 &= 3\end{aligned}$

Latihan Soal

Nah, sekarang latihan soal.

1. Hitunglah!
   • $\displaystyle\lim_{x\to3}(x+3)(x+2)^2$
   • $\displaystyle\lim_{x\to3}\sqrt{(x+3)^2-(x+2)^2}$
2. Jika diketahui $\displaystyle\lim_{x\to a}f(x)=1$ dan $\displaystyle\lim_{x\to a}g(x)=3$. Tentukan nilai dari
   • $\displaystyle \lim_{x\to a} \left[f(x) + 2g(x)\right]$
     
   • $\displaystyle \lim_{x\to a} \left[3f(x) - g(x)\right]^5$
     
   • $\displaystyle \lim_{x\to a} \sqrt{f(x)^2 + g(x)^2}$

Selamat Belajar!

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