LIMIT FUNGSI e

Salah satu permasalahan dalam limit adalah mengenai limit e. Untuk rumus dasar limit e bisa berikut ini,
limx→∞(1+1n)n=e$\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{n}\right)}^n=e$ 
limx→∞(1−1n)−n=e$\underset{x\to \mathrm{\infty }}{lim}{\left(1-\frac{1}{n}\right)}^{-n}=e$ 
limx→0(1+n)1n=e$\underset{x\to 0}{lim}{\left(1+n\right)}^{\frac{1}{n}}=e$ 

Inti dari penyelesaian soal dengan limit e tersebut adalah membuat bentuk umum menjadi bentuk e tersebut. Untuk lebih memudahkan pemahaman, coba perhatikan contoh soal dan pembahasan tentang limit e ini.

1. 

limx→∞(1+f(x))g(x)=limx→∞(1+1h(x))g(x)$\underset{x\to \infty }{\lim }{\left(1+f\left(x\right)\right)}^{g\left(x\right)}=\underset{x\to \infty }{\lim }{\left(1+\frac{1}{h\left(x\right)}\right)}^{g\left(x\right)}$ dengan h(x)=1f(x)$h\left(x\right)=\frac{1}{f\left(x\right)}$

=limx→∞{(1+1h(x))h(x)}g(x)h(x)$=\underset{x\to \infty }{\lim }{\left\{{\left(1+\frac{1}{h\left(x\right)}\right)}^{h\left(x\right)}\right\}}^{\frac{g\left(x\right)}{h\left(x\right)}}$

=limx→∞{e}g(x)h(x)$=\underset{x\to \infty }{\lim }{\left\{e\right\}}^{\frac{g\left(x\right)}{h\left(x\right)}}$

=(e)limx→∞g(x)h(x)$={\left(e\right)}^{\underset{x\to \infty }{\lim }\frac{g\left(x\right)}{h\left(x\right)}}$ kembalikan h(x)=1f(x)$h\left(x\right)=\frac{1}{f\left(x\right)}$

=(e)limx→∞f(x)×g(x)

6.

2. Tentukan limit dari : limx→∞(1+xx−1)6x$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{1+x}{x-1}\right)}^{6x}$  Untuk menyelesaikan ini jadikan dalam bentuk umum seperti rumus di atas, limx→∞(1+xx−1)6x$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{1+x}{x-1}\right)}^{6x}$ limx→∞(1+x−1+1x−1)6x$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{1+x-1+1}{x-1}\right)}^{6x}$ limx→∞(x−1+2x−1)6x$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x-1+2}{x-1}\right)}^{6x}$ limx→∞(1+2x−1)6x$\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{2}{x-1}\right)}^{6x}$ limx→∞(1+1x−12)6x$\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{x-1}{2}}{)}^{6x}$ Jadi n kita disini adalah x−12$\frac{x-1}{2}$ Untuk pangkat langsung kita kalikan dengan n/n sehingga diperoleh :  6x.x−12x−12$Ambilsebuah$x−12$6x.\frac{\frac{x-1}{2}}{\frac{x-1}{2}\$Ambilsebuah\$\frac{x-1}{2}}$ untuk membuat fungsi menjadi e. limx→∞(1+1x−12)x−12=e contoh lain.Tentukan hasil dari a. limx→∞(1+8x)x6+2x4√−x6−2x4√$\underset{x\to \infty }{\lim }{\left(1+\frac{8}{x}\right)}^{\sqrt{x^6+2x^4}-\sqrt{x^6-2x^4}}$ c. limx→∞(x−1x+2)(5x−1)22x+5$\underset{x\to \infty }{\lim }{\left(\frac{x-1}{x+2}\right)}^{\frac{{\left(5x-1\right)}^2}{2x+5}}$ b. limx→∞(1−12x)3x2+6x−112x+5$\underset{x\to \infty }{\lim }{\left(1-\frac{1}{2x}\right)}^{\frac{{3x}^2+6x-11}{2x+5}}$ d. limx→∞(3x+13x−5)2x5+4x+3(2x−1)2(x−2)2 jawab : a. limx→∞(1+8x)x6+2x4√−x6−2x4√=(e)limx→∞{8x(x6+2x4√−x6−2x4√)}$\underset{x\to \infty }{\lim }{\left(1+\frac{8}{x}\right)}^{\sqrt{x^6+2x^4}-\sqrt{x^6-2x^4}}={\left(e\right)}^{\underset{x\to \infty }{\lim }\left\{\frac{8}{x}(\sqrt{x^6+2x^4}-\sqrt{x^6-2x^4})\right\}}$ =(e)limx→∞8{(x6+2x4)−(x6−2x4)}x(x6+2x4√+x6−2x4√)$={\left(e\right)}^{\underset{x\to \infty }{\lim }\frac{8\left\{\left(x^6+2x^4\right)-\left(x^6-2x^4\right)\right\}}{x(\sqrt{x^6+2x^4}+\sqrt{x^6-2x^4})}}$   =(e)limx→∞8{4x4}x(x6+2x4√+x6−2x4√)$={\left(e\right)}^{\underset{x\to \infty }{\lim }\frac{8\left\{4x^4\right\}}{x(\sqrt{x^6+2x^4}+\sqrt{x^6-2x^4})}}$   = (e)8(4)1(1√+1√)${\left(e\right)}^{\frac{8\left(4\right)}{1\left(\sqrt{1}+\sqrt{1}\right)}}$ =e16$=e^{16}$ b. limx→∞(1−12x)3x2+6x−112x+5=(e)limx→∞{(−12x)(3x2+6x−112x+5)}$\underset{x\to \infty }{\lim }{\left(1-\frac{1}{2x}\right)}^{\frac{{3x}^2+6x-11}{2x+5}}={\left(e\right)}^{\underset{x\to \infty }{\lim }\left\{\left(-\frac{1}{2x}\right)\left(\frac{{3x}^2+6x-11}{2x+5}\right)\right\}}$ =(e)−12×32$={\left(e\right)}^{-\frac{1}{2}\times \frac{3}{2}}$   =e−34$=e^{-\frac{3}{4}}$ c. limx→∞(x−1x+2)(5x−1)22x+5=limx→∞(1−3x+2)(5x−1)22x+5$\underset{x\to \infty }{\lim }{\left(\frac{x-1}{x+2}\right)}^{\frac{{\left(5x-1\right)}^2}{2x+5}}=\underset{x\to \infty }{\lim }{\left(1-\frac{3}{x+2}\right)}^{\frac{{\left(5x-1\right)}^2}{2x+5}}$ =(e)limx→∞{(−3x+2)((5x−1)22x+5)}$={\left(e\right)}^{\underset{x\to \infty }{\lim }\left\{\left(-\frac{3}{x+2}\right)\left(\frac{{\left(5x-1\right)}^2}{2x+5}\right)\right\}}$ =e−31×522$=e^{-\frac{3}{1}\times \frac{5^2}{2}}$ =e−752$=e^{-\frac{75}{2}}$   =1e37e√$=\frac{1}{e^{37}\sqrt{e}}$ d. limx→∞(3x+13x−5)2x5+4x+3(2x−1)2(x−2)2=limx→∞(1+63x−5)2x5+4x+3(2x−1)2(x−2)2$\underset{x\to \infty }{\lim }{\left(\frac{3x+1}{3x-5}\right)}^{\frac{{2x}^5+4x+3}{{(2x-1)}^2{(x-2)}^2}}=\underset{x\to \infty }{\lim }{\left(1+\frac{6}{3x-5}\right)}^{\frac{{2x}^5+4x+3}{{(2x-1)}^2{(x-2)}^2}}$ =(e)limx→∞{(63x−5)(2x5+4x+3(2x−1)2(x−2)2)}$={\left(e\right)}^{\underset{x\to \infty }{\lim }\left\{\left(\frac{6}{3x-5}\right)\left(\frac{{2x}^5+4x+3}{{(2x-1)}^2{(x-2)}^2}\right)\right\}}$ =e63×222.12$=e^{\frac{6}{3}\times \frac{2}{2^2.1^2}}$ =e