Tag: Ekspansi Newton

Perhatikan bahwa
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{2x}})}^{5x}=\underset{x\to \mathrm{\infty }}{lim}{\left({(1+{\displaystyle \frac{1}{2x}})}^{2x}\right)}^{\frac{5}{2}}.}$$Karena $x\to \mathrm{\infty }$, haruslah $2x\to \mathrm{\infty }$ sehingga dapat ditulis ${\left({\displaystyle \underset{2x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{2x}})}^{2x}}\right)}^{\frac{5}{2}}=e^{\frac{5}{2}}.$
Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{2x}})}^{5x}=e^{\frac{5}{2}}=e^2\sqrt{e}}}}.$

Perhatikan bahwa
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{x+5}{x+3}}\right)}^{x+6}}& =\underset{x\to \mathrm{\infty }}{lim}{({\displaystyle \frac{x+3}{x+3}}+{\displaystyle \frac{2}{x+3}})}^{x+6}\\ & =\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{{\displaystyle \frac{x+3}{2}}}})}^{\frac{x+3}{2}\cdot \frac{x+6}{x+3}\cdot 2}\end{array}$$Karena $x\to \mathrm{\infty },$ haruslah ${\displaystyle \frac{x+3}{2}}\to \mathrm{\infty }$ sehingga selanjutnya dapat ditulis
$${\displaystyle {\underset{\frac{x+3}{2}\to \mathrm{\infty }}{lim}(1+{\displaystyle \frac{1}{{\displaystyle \frac{x+3}{2}}}})}^{\frac{x+3}{2}\cdot \frac{x+6}{x+3}\cdot 2}.}$$Bagian yang ditandai warna merah di atas membentuk limit euler. Tinjau bahwa bentuk ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}({\displaystyle \frac{x+6}{x+3}}\cdot 2)=1\times 2=2.}$
Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{x+5}{x+3}}\right)}^{x+6}=e^2}}}$

Perhatikan bahwa
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1-{\displaystyle \frac{2}{x}})}^x={\left(\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{-{\displaystyle \frac{x}{2}}}})}^{-\frac{x}{2}}\right)}^{-2}.}$$Karena $x\to \mathrm{\infty }$, haruslah ${\displaystyle \frac{x}{2}}\to \mathrm{\infty }$ sehingga diperoleh
${\left({\displaystyle \underset{x\to \frac{\pi }{2}}{lim}{(1+{\displaystyle \frac{1}{-{\displaystyle \frac{x}{2}}}})}^{-\frac{x}{2}}}\right)}^{-2}=e^{-2}.$
Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{-2}}}$

Perhatikan bahwa
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left(\sqrt{1+{\displaystyle \frac{1}{x}}}\right)}^{3x}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{x}})}^{\frac{3}{2}x}& & (\sqrt{a}=a^{\frac{1}{2}})\\ & ={\left(\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{x}})}^x\right)}^{\frac{3}{2}}\\ & =e^{\frac{3}{2}}=e\sqrt{e}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e\sqrt{e}}}.$

Karena $x\to 0$, haruslah $-{\displaystyle \frac{1}{3}}x\to 0$ sehingga dapat ditulis
$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{(1-{\displaystyle \frac{x}{3}})}^{\frac{1}{4x}}}& ={\left[\underset{-\frac{1}{3}x\to 0}{lim}{(1+(-{\displaystyle \frac{1}{3}}x))}^{\frac{1}{(-\frac{1}{3}x)}}\right]}^{-\frac{1}{12}}\\ & =e^{-\frac{1}{12}}={\displaystyle \frac{1}{\sqrt[12]{e}}}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle {\displaystyle \frac{1}{\sqrt[12]{e}}}}}.$

Perhatikan bahwa $1+4x+4x^2$ dapat difaktorkan menjadi $(1+2x{)}^2$ sehingga 
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(1+4x+4x^2{)}^{\frac{3}{x}}}& =\underset{x\to \mathrm{\infty }}{lim}(1+2x{)}^{\frac{6}{x}}\\ & ={[\underset{x\to \mathrm{\infty }}{lim}(1+2x{)}^{\frac{1}{2x}}]}^{12}\\ & ={\left[{\underset{2x\to \mathrm{\infty }}{lim}(1+2x)}^{\frac{1}{2x}}\right]}^{12}& & (x\to \mathrm{\infty }⟹2x\to \mathrm{\infty })\\ & =e^{12}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{12}}}.$

Perhatikan bawah
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1-{\displaystyle \frac{1}{3x}})}^{12x}}\\ & ={\left[\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{-3x}})}^{-3x}\right]}^{-4}\\ & ={\left[\underset{-3x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{-3x}})}^{-3x}\right]}^{-4}=e^{-4}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{-4}}}.$

Perhatikan bahwa
$$\text{Missing begin{aligned} or extra end{aligned}}$$Karena $x\to \mathrm{\infty }$, haruslah ${\displaystyle \frac{x^2+4}{3x+5}}\to \mathrm{\infty }$ (pangkat terbesar ada di pembilang, berarti nilai limitnya tak hingga) sehingga selanjutnya dapat ditulis
$${\displaystyle {\underset{\frac{x^2+4}{3x+5}\to \mathrm{\infty }}{lim}(1+{\displaystyle \frac{1}{{\displaystyle \frac{x^2+4}{3x+5}}}})}^{\frac{x^2+4}{3x+5}\cdot \frac{3x+5}{x^2+4}\cdot 2x}.}$$Bagian yang ditandai warna merah di atas membentuk limit euler. Tinjau bahwa bentuk
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}({\displaystyle \frac{3x+5}{x^2+4}}\cdot 2x)}& =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{6x^2+10x}{x^2+4}}\\ & =\underset{x\to \mathrm{\infty }}{lim}={\displaystyle \frac{6}{1}}=6.\end{array}$$Dengan demikian,
$\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{3x+5}{x^2+4}})}^{2x}=e^6}}}.$

Perhatikan bahwa
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{3x+1}{3x+5}}\right)}^{x^2+3}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{({\displaystyle \frac{3x+5}{3x+5}}+{\displaystyle \frac{-4}{3x+5}})}^{x^2+3}\\ & =\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{{\displaystyle \frac{3x+5}{-4}}}})}^{\frac{3x+5}{-4}\cdot \frac{-4}{3x+5}\cdot (x^2+3)}.\end{array}$$Karena $x\to \mathrm{\infty }$, haruslah ${\displaystyle \frac{3x+5}{-4}}\to \mathrm{\infty }$ (pangkat terbesar ada di pembilang, berarti nilai limitnya takhingga) sehingga selanjutnya dapat ditulis
$${\displaystyle {\underset{\frac{3x+5}{-4}\to \mathrm{\infty }}{lim}(1+{\displaystyle \frac{1}{{\displaystyle \frac{3x+5}{-4}}}})}^{\frac{3x+5}{-4}\cdot \frac{-4}{3x+5}\cdot (x^2+3)}.}$$Bagian yang ditandai warna merah di atas membentuk limit euler. Tinjau bahwa bentuk
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}({\displaystyle \frac{-4}{3x+5}}\cdot (x^2+3))=\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-4x^2-12}{3x+5}}=-\mathrm{\infty }.}$$Dengan demikian, $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{3x+1}{3x+5}}\right)}^{x^2+3}=e^{-\mathrm{\infty }}=0}}}.$

Alternatif 1: Menggunakan ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{n}})}^n=e}$
Misalkan $x={\displaystyle \frac{1}{y}}$. Karena $x\to 0$, maka $y\to \mathrm{\infty }.$ Dengan demikian, dapat kita tulis
$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}(1+3x{)}^{\frac{2}{x}}}& =\underset{y\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{3}{y}})}^{2y}\\ & ={\left(\underset{y\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{{\displaystyle \frac{y}{3}}}})}^{\frac{y}{3}}\right)}^6& & (2y={\displaystyle \frac{y}{3}}\times 6).\end{array}$$Karena $y\to \mathrm{\infty }$, maka ${\displaystyle \frac{y}{3}}\to \mathrm{\infty }$ sehingga
${\left[{\displaystyle \underset{\frac{y}{3}\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{{\displaystyle \frac{y}{3}}}})}^{\frac{y}{3}}}\right]}^6=e^6.$
Alternatif 2: Menggunakan ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(1+n{)}^{\frac{1}{n}}=e.}$
Karena $x\to 0$, maka $3x\to 0$ sehingga dapat ditulis
$\begin{array}{rl}& {\displaystyle \underset{x\to 0}{lim}(1+3x{)}^{\frac{2}{x}}}\\ & ={\left[\underset{3x\to 0}{lim}(1+3x{)}^{\frac{1}{3x}}\right]}^6=e^6.\end{array}$
Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^6}}.$

Perhatikan bahwa
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{x-1}{x+1}}\right)}^{3x-2}}& =\underset{x\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{(x+1)-2}{x+1}}\right)}^{3x-2}\\ & =\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{-2}{x+1}})}^{3x-2}.\end{array}$$Jika berturut-turut dimisalkan $f(x)={\displaystyle \frac{-2}{x+1}}$ dan $g(x)=3x-2$, maka diperoleh
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-2}{x+1}}=0}$
dan
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}g(x)=\underset{x\to \mathrm{\infty }}{lim}(3x-2)=\mathrm{\infty }.}$
Berdasarkan teorema 1 Limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{-2}{x+1}})}^{3x-2}}& =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-2}{x+1}}\cdot (3x-2)}}\\ & =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-6x+4}{x+1}}}}\\ & =e^{-6}.\end{array}$$Jadi, nilai dari limit tersebut adalah $\boxed{{\displaystyle e^{-6}}}$

Perhatikan bahwa 
$${\displaystyle \underset{x\to 1}{lim}{x}^{\frac{x}{x^2-3x+2}}=\underset{x\to 1}{lim}(1+(x-1){)}^{\frac{x}{(x-1)(x-2)}}.}$$Jika berturut-turut dimisalkan $f(x)=x-1$ dan $g(x)={\displaystyle \frac{x}{(x-1)(x-2)}},$ maka diperoleh
${\displaystyle \underset{x\to 1}{lim}f(x)=\underset{x\to 1}{lim}(x-1)=0}$
dan
$${\displaystyle \underset{x\to 1}{lim}g(x)=\{\begin{array}{l}{\displaystyle \underset{x\to 1^{+}}{lim}{\displaystyle \frac{x}{(x-1)(x-2)}}=-\mathrm{\infty }}\\ {\displaystyle \underset{x\to 1^{-}}{lim}{\displaystyle \frac{x}{(x-1)(x-2)}}=\mathrm{\infty }.}\end{array}}$$Berdasarkan teorema 1 Limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}{(1+(x-1))}^{\frac{x}{(x-1)(x-2)}}}& =e^{{\displaystyle \underset{x\to 1}{lim}\cancel{(x-1)}\cdot \frac{x}{\cancel{(x-1)}(x-2)}}}\\ & =e^{{\displaystyle \underset{x\to 1}{lim}\frac{x}{x-2}}}=e^{\frac{1}{1-2}}\\ & =e^{-1}.\end{array}$$Jadi, nilai dari limit tersebut adalah $\boxed{{\displaystyle e^{-1}}}.$

Misalkan $f(x)={\displaystyle \frac{8}{x}}$ dan $g(x)=\sqrt{x^6+2x^4}-\sqrt{x^6-2x^4}$ sehingga
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{8}{x}}=0}$
dan
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}g(x)}& =\underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^6+2x^4}-\sqrt{x^6-2x^4})\times {\displaystyle \frac{\sqrt{x^6+2x^4}+\sqrt{x^6-2x^4}}{\sqrt{x^6+2x^4}+\sqrt{x^6-2x^4}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(x^6+2x^4)-(x^6-2x^4)}{\sqrt{x^6+2x^4}+\sqrt{x^6-2x^4}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{4x^4}{\sqrt{x^6+2x^4}+\sqrt{x^6-2x^4}}}=\mathrm{\infty }.\end{array}$$Catatan: Limit di atas bernilai $\mathrm{\infty }$ karena bentuk terakhir menunjukkan bahwa pembilang memiliki derajat yang lebih besar dari penyebut.
Dengan menggunakan teorema 1 Limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{8}{x}})}^{\sqrt{x^6+2x^4}-\sqrt{x^6-2x^4}}}& =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)g(x)}}\\ & =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{8}{x}\cdot (\sqrt{x^6+2x^4}-\sqrt{x^6-2x^4})}}\\ & =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{8(4x^4)}{x(\sqrt{x^6+2x^4}+\sqrt{x^6-2x^4})}}}}\\ & =e^{{\displaystyle 32\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x^3}{\sqrt{x^6+2x^4}+\sqrt{x^6-2x^4}}}}}\\ & =e^{32\left(\frac{1}{\sqrt{1+0}+\sqrt{1-0}}\right)}\\ & =e^{16}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{16}}}.$

Misalkan $f(x)=-{\displaystyle \frac{1}{2x}}$ dan $g(x)={\displaystyle \frac{3x^2+6x-11}{2x+5}}$ sehingga
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}-{\displaystyle \frac{1}{2x}}=0}$
dan
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}g(x)=\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{3x^2+6x-11}{2x+5}}}$ $=\mathrm{\infty }.$
Dengan menggunakan teorema 1 Limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1-{\displaystyle \frac{1}{2x}})}^{\frac{3x^2+6x-11}{2x+5}}}& =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)g(x)}}\\ & =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}-{\displaystyle \frac{1}{2x}}\cdot \frac{3x^2+6x-11}{2x+5}}}\\ & =e^{-{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{3x^2+6x-11}{4x^2+10x}}}\\ & =e^{-\frac{3}{4}}\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{-\frac{3}{4}}}}.$

Perhatikan bahwa
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{x-1}{x+2}}\right)}^{\frac{(5x-1{)}^2}{2x+5}}=\underset{x\to \mathrm{\infty }}{lim}{(1-{\displaystyle \frac{3}{x+2}})}^{\frac{(5x-1{)}^2}{2x+5}}.}$$Misalkan $f(x)=-{\displaystyle \frac{3}{x+2}}$ dan $g(x)={\displaystyle \frac{(5x-1{)}^2}{2x+5}}$ sehingga
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}-{\displaystyle \frac{3}{x+2}}=0}$
dan
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}g(x)=\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(5x-1{)}^2}{2x+5}}=\mathrm{\infty }.}$
Dengan menggunakan teorema 1 Limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1-{\displaystyle \frac{3}{x+2}})}^{\frac{(5x-1{)}^2}{2x+5}}}}& =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)g(x)}}\\ & =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}-{\displaystyle \frac{3}{x+2}}\cdot \frac{(5x-1{)}^2}{2x+5}}}\\ & =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{-75x^2+30x-3}{2x^2+9x+10}}}\\ & =e^{-\frac{75}{2}}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{-\frac{75}{2}}}}$

Perhatikan bahwa
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{3x+1}{3x-5}}\right)}^{\frac{2x^5+4x+3}{(2x-1{)}^2(x-2{)}^2}}=\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{6}{3x-5}})}^{\frac{2x^5+4x+3}{(2x-1{)}^2(x-2{)}^2}}.}$$Misalkan $f(x)={\displaystyle \frac{6}{3x-5}}$ dan $$g(x)={\displaystyle \frac{2x^5+4x+3}{(2x-1{)}^2(x-2{)}^2}}={\displaystyle \frac{2x^5+4x+3}{4x^4+\cdots +\cdots }}$$ sehingga
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)=\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{6}{3x-5}}=0}$
dan
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}g(x)=\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2x^5+4x+3}{4x^4+\cdots +\cdots }}}$ $=\mathrm{\infty }.$
Dengan menggunakan teorema 1 Limit Euler, diperoleh
$$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{6}{3x-5}})}^{\frac{2x^5+4x+3}{(2x-1{)}^2(x-2{)}^2}}& =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)g(x)}}\\ & =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{6}{3x-5}}\cdot \frac{2x^5+4x+3}{4x^4+\cdots +\cdots }}}\\ & =e^{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{12x^5+24x+18}{12x^5+\cdots +\cdots }}}\\ & =e^{\frac{12}{12}}=e\end{array}$$Catatan: Karena nilai limit tak hingga pada fungsi rasional polinomial dapat serta merta ditentukan dengan hanya memperhatikan koefisien variabel berpangkat tertinggi pada pembilang dan penyebut, maka variabel berderajat lebih rendah dapat diabaikan.
Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e}}$

Misalkan $f(x)={\displaystyle \frac{x}{2}}$ dan $g(x)={\displaystyle \frac{x}{{\sin }^{2}3x}}$ sehingga
${\displaystyle \underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}{\displaystyle \frac{x}{2}}=0}$
dan
${\displaystyle \underset{x\to 0}{lim}g(x)=\underset{x\to 0}{lim}{\displaystyle \frac{x}{{\sin }^{2}3x}}=\pm \mathrm{\infty }.}$
Dengan menggunakan teorema 1 limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{(1+{\displaystyle \frac{x}{2}})}^{\frac{x}{{\sin }^{2}3x}}}& =e^{{\displaystyle \underset{x\to 0}{lim}f(x)g(x)}}\\ & =e^{{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{2}}\cdot \frac{x}{{\sin }^{2}3x}}}\\ & =e^{\frac{1}{2}{\displaystyle \underset{x\to 0}{lim}\frac{x}{\sin 3x}\cdot \frac{x}{\sin 3x}}}\\ & =e^{\frac{1}{2}\cdot \frac{1}{3}\cdot \frac{1}{3}}=e^{\frac{1}{18}}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{\frac{1}{18}}}}$

Misalkan $f(x)=-2x$ dan $g(x)={\displaystyle \frac{1}{\sin 3x}}$ sehingga
${\displaystyle \underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}-2x=0}$
dan
${\displaystyle \underset{x\to 0}{lim}g(x)=\underset{x\to 0}{lim}{\displaystyle \frac{1}{\sin 3x}}=\pm \mathrm{\infty }.}$
Dengan menggunakan teorema 1 limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}(1-2x{)}^{\frac{1}{\sin 3x}}}& =e^{{\displaystyle \underset{x\to 0}{lim}f(x)g(x)}}\\ & =e^{{\displaystyle \underset{x\to 0}{lim}(-2x)\cdot \frac{1}{\sin 3x}}}\\ & =e^{-2{\displaystyle \underset{x\to 0}{lim}\frac{x}{\sin 3x}}}\\ & =e^{-2\cdot \frac{1}{3}}=e^{-\frac{2}{3}}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{-\frac{2}{3}}}}$

Karena $\sqrt{a}=a^{\frac{1}{2}}$, bentuk limit di atas dapat ditulis sebagai
${\displaystyle \underset{x\to 0}{lim}(1+3x{)}^{\frac{1}{2}\cdot \frac{\sin 6x}{1-\cos 4x}}.}$
Misalkan $f(x)=3x$ dan $g(x)=\frac{1}{2}\cdot {\displaystyle \frac{\sin 6x}{1-\cos 4x}}$ sehingga
${\displaystyle \underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}3x=0}$
dan
${\displaystyle \underset{x\to 0}{lim}g(x)=\underset{x\to 0}{lim}\frac{1}{2}\cdot {\displaystyle \frac{\sin 6x}{1-\cos 4x}}}$ $=\pm \mathrm{\infty }.$
Dengan menggunakan teorema 1 limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}(1+3x{)}^{\frac{1}{2}\cdot \frac{\sin 6x}{1-\cos 4x}}}& =e^{{\displaystyle \underset{x\to 0}{lim}f(x)g(x)}}\\ & =e^{{\displaystyle \underset{x\to 0}{lim}(3x)\cdot \frac{1}{2}\cdot \frac{\sin 6x}{1-\cos 4x}}}\\ & =e^{\frac{1}{2}{\displaystyle \underset{x\to 0}{lim}\frac{3x\sin 6x}{1-(1-2{\sin }^{2}2x)}}}\\ & =e^{\frac{1}{2}{\displaystyle \underset{x\to 0}{lim}\frac{3x\sin 6x}{2{\sin }^{2}2x}}}\\ & =e^{\frac{3}{4}{\displaystyle \underset{x\to 0}{lim}\frac{x}{\sin 2x}\cdot \frac{\sin 6x}{\sin 2x}}}\\ & =e^{\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{6}{2}}=e^{\frac{9}{8}}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{\frac{9}{8}}}}.$

Perhatikan bahwa bentuk $(1-10x+25x^2)$ dapat difaktorkan menjadi $(1-5x{)}^2$ sehingga bentuk limitnya dapat ditulis menjadi ${\displaystyle \underset{x\to 0}{lim}(1-5x{)}^{2\cdot \frac{\cos 3x}{\sin 6x}}.}$
Misalkan $f(x)=-5x$ dan $g(x)=2\cdot {\displaystyle \frac{\cos 3x}{\sin 6x}}$ sehingga
${\displaystyle \underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}(-5x)=0}$
dan
${\displaystyle \underset{x\to 0}{lim}g(x)=\underset{x\to 0}{lim}2\cdot {\displaystyle \frac{\cos 3x}{\sin 6x}}=\pm \mathrm{\infty }.}$
Dengan menggunakan teorema 1 limit Euler, diperoleh
$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}(1-5x{)}^{2\cdot \frac{\cos 3x}{\sin 6x}}}& =e^{{\displaystyle \underset{x\to 0}{lim}f(x)g(x)}}\\ & =e^{{\displaystyle \underset{x\to 0}{lim}(-5x)(2\cdot {\displaystyle \frac{\cos 3x}{\sin 6x}})}}\\ & =e^{-10{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{\sin 6x}}\cdot \cos 3x}}\\ & =e^{-10\cdot \frac{1}{6}\cdot 1}\\ & =e^{-\frac{5}{3}}.\end{array}$$Jadi, nilai dari limit di atas adalah $\boxed{{\displaystyle e^{-\frac{5}{3}}}}$

Berdasarkan rumus penjumlahan $n$ bilangan asli dan $n$ kuadrat bilangan asli, kita peroleh
$$\begin{array}{rl}2n(1+2+3+\cdots +n)& =2n\cdot {\displaystyle \frac{n(n+1)}{2}}\\ & =n^2(n+1)\\ 3(1^2+2^2+3^2+\cdots +n^2)& =3\cdot {\displaystyle \frac{n(n+1)(2n+1)}{6}}\\ & ={\displaystyle \frac{n(n+1)(2n+1)}{2}}.\end{array}$$Untuk itu, kita akan mendapatkan
$$\begin{array}{rl}& {\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\left[{\displaystyle \frac{2n(1+2+3+\cdots +n)}{3(1^2+2^2+3^2+\cdots +n^2)}}\right]}^n}\\ & =\underset{n\to \mathrm{\infty }}{lim}{[n^2(n+1)÷{\displaystyle \frac{n(n+1)(2n+1)}{2}}]}^n\\ & =\underset{n\to \mathrm{\infty }}{lim}{\left[{\displaystyle \frac{2n}{2n+1}}\right]}^n\\ & =\underset{n\to \mathrm{\infty }}{lim}{\left[{\displaystyle \frac{2n+1-1}{2n+1}}\right]}^n\\ & =\underset{n\to \mathrm{\infty }}{lim}{[1-{\displaystyle \frac{1}{2n+1}}]}^n\\ & =e^{{\displaystyle \underset{n\to \mathrm{\infty }}{lim}(-{\displaystyle \frac{1}{2n+1}}\cdot n)}}\\ & =e^{-\frac{1}{2}}={\displaystyle \frac{1}{\sqrt{e}}}.\end{array}$$Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\left[{\displaystyle \frac{2n(1+2+3+\cdots +n)}{3(1^2+2^2+3^2+\cdots +n^2)}}\right]}^n={\displaystyle \frac{1}{\sqrt{e}}}}}}.$$