$\lim_{x \to0} \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)$

[alpercay]

$\lim_{x \to0}  \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)$ limitini bulunuz.

[Metin Can Aydemir]

Payda eşitleyelim ve $\sin^2{x}=\frac{1}{2}(1-\cos{2x})=\frac{1}{2}\left(\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!}+\cdots\right)$ olduğunu kullanalım. $$\lim\limits_{x\to 0}\left(\frac{x^2-\sin^2{x}}{x^2\sin^2{x}}\right)=\lim\limits_{x\to 0}\left(\frac{x^2-\frac{1}{2}\left(\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!}+\cdots\right)}{x^2\frac{1}{2}\left(\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!}+\cdots\right)}\right)$$ $$=\lim\limits_{x\to 0}\left(\frac{2x^2-\left(\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!}+\cdots\right)}{x^2\left(\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!}+\cdots\right)}\right)$$ $$=\lim\limits_{x\to 0}\left(\frac{\frac{(2x)^4}{4!}-\frac{(2x)^6}{6!}+\cdots}{\frac{(2x)^2x^2}{2!}-\frac{(2x)^4x^2}{4!}+\cdots}\right)=\frac{\frac{2^4}{4!}}{\frac{2^2}{2!}}=\frac{1}{3}.$$

[alpercay]

Payda eşitleyip pay ve paydayı $x$ ile genişletelim:

$\lim_{x \to0}  \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)=\lim_{x \to0}\dfrac{x(x-\sin x)(x+\sin x)}{x^3\sin^2x}$

$=\lim_{x \to0}\left(\dfrac{x}{\sin x}\right)\cdot\lim_{x \to0}\dfrac{x+\sin x}{\sin x}\cdot\lim_{x \to0}\dfrac{1-\cos x}{3x^2}$

$=1.\lim_{x \to0}\dfrac{1+\cos x}{\cos x}\cdot\lim_{x \to0}\dfrac{1-\cos x}{3x^2}=1.2.\lim_{x \to0}\dfrac{\sin x}{6x}=\dfrac 13$

[alpercay]

$\lim_{ x \to 0}\left(\dfrac {1}{\sin^2x}-\dfrac1{x^2}\right)=L$ olsun.

$x=2y$ dönüşümü yapalım:

$\lim_{ y \to 0}\left(\dfrac {1}{\sin^22y}-\dfrac1{4y^2}\right)=\lim_{ y \to 0}\left(\dfrac1{4\sin^2y\cdot\cos^2y}-\dfrac1{4y^2}\right)=L$

$=\dfrac14\lim_{ y \to 0}\left(\dfrac1{\cos^2{y}}+\dfrac1{\sin^2{y}}-\dfrac1{y^2}\right)=\dfrac14+\dfrac{L}{4}=L$

$L=\dfrac13$

[Metin Can Aydemir]

$\lim_{ x \to 0}\left(\dfrac {1}{\sin^2x}-\dfrac1{x^2}\right)=L$ olsun.

$x=2y$ dönüşümü yapalım:

$\lim_{ y \to 0}\left(\dfrac {1}{\sin^22y}-\dfrac1{4y^2}\right)=\lim_{ y \to 0}\left(\dfrac1{4\sin^2y\cdot\cos^2y}-\dfrac1{4y^2}\right)=L$

$=\dfrac14\lim_{ y \to 0}\left(\dfrac1{\cos^2{y}}+\dfrac1{\sin^2{y}}-\dfrac1{y^2}\right)=\dfrac14+\dfrac{L}{4}=L$

$L=\dfrac13$

Ya limit $+\infty$ ise?

[alpercay]

Haklısın, limitin sonlu bir sayıya eşit olduğunu kabül ederek çözüm yapıldı. Yakınsaklık da gösterilmeliydi.