Giới hạn cơ bản: $\mathop {\lim }\limits_{x \to 0}\frac{\ln(1+x)}{x}=1$
 Ta có:
     $\mathop {\lim }\limits_{x \to +\infty}x^2\ln\frac{x^2-1}{x^2+1}$
$=\mathop {\lim }\limits_{x \to +\infty}x^2\ln(1-\frac{2}{x^2+1})$
$=\mathop {\lim }\limits_{x \to +\infty}\frac{\ln(1-\frac{2}{x^2+1})}{-\frac{2}{x^2+1}}.\frac{-2x^2}{x^2+1}=-2$

Giới hạn cơ bản: $\mathop {\lim }\limits_{x \to 0}\frac{\sin x}{x}=1$
Ta có:
    $\mathop {\lim }\limits_{x \to \infty}x\sin\frac{1}{x}$
$=\mathop {\lim }\limits_{x \to \infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}=1$

Giới hạn cơ bản: $\mathop {\lim }\limits_{x \to 0}\frac{\sin x}{x}=\mathop {\lim }\limits_{x \to 0}\frac{x}{\sin x}=1$
Ta có:
     $\mathop {\lim }\limits_{x \to 0^+}\frac{x}{\sqrt{1-\cos x}}$
$=\mathop {\lim }\limits_{x \to 0^+}\frac{x}{\sqrt2\sin\frac{x}{2}}$
$=\mathop {\lim }\limits_{x \to 0^+}\frac{\sqrt2.\frac{x}{2}}{\sin\frac{x}{2}}=\sqrt2$
Mặt khác:
     $\mathop {\lim }\limits_{x \to 0^-}\frac{x}{\sqrt{1-\cos x}}$
$=\mathop {\lim }\limits_{x \to 0^-}\frac{x}{-\sqrt2\sin\frac{x}{2}}$
$=\mathop {\lim }\limits_{x \to 0^-}\frac{-\sqrt2.\frac{x}{2}}{\sin\frac{x}{2}}=-\sqrt2$
Suy ra không tồn tại $\mathop {\lim }\limits_{x \to 0}\frac{x}{\sqrt{1-\cos x}}$

Giới hạn cơ bản: $\mathop {\lim }\limits_{x \to 0}\frac{\ln(1+x)}{x}=1$
Ta có:
$\mathop {\lim }\limits_{x \to +\infty}\left(1+\frac{2}{x}\right)^x=e^{\mathop {\lim }\limits_{x \to +\infty}x\ln(1+\frac{2}{x})}$
Mà ta có:
   $\mathop {\lim }\limits_{x \to +\infty}x\ln(1+\frac{2}{x})$
$=\mathop {\lim }\limits_{x \to 1}\frac{\ln(1+\frac{2}{x})}{\frac{2}{x}}.2=2$
Suy ra:
$\mathop {\lim }\limits_{x \to +\infty}\left(1+\frac{2}{x}\right)^x=e^2$

Bài 4:
Ta có:
      $\left\{\begin{array}{l}u_5-u_3=72\\u_3+u_2=36\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}u_1(q^4-q^2)=72\\u_1(q^2+q)=36\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}q^4-q^2=2(q^2+q)\\u_1q(q+1)=36\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}q=2\\u_1=6\end{array}\right.$, vì $q(q+1)\ne 0$
Lại có:
      $S_n=u_1.\frac{1-q^n}{1-q}$
$\Leftrightarrow 1530=6.\frac{1-2^n}{1-2}$
$\Leftrightarrow 2^n=256\Leftrightarrow n=8$

Bài 3:
Gọi 3 số hạng cần tìm là: $\frac{a}{q};a;aq$, với $a,q\ne0$
Từ giả thiết suy ra: $\frac{a}{q}.a.aq=216\Leftrightarrow a^3=216\Leftrightarrow a=6$
Lại có: $\frac{a}{q}+a+aq=19$
         $\Leftrightarrow q+\frac{1}{q}=\frac{13}{6}$
         $\Leftrightarrow 6a^2-13q+6=0$
         $\Leftrightarrow \left[\begin{array}{l}q=\frac{3}{2}\\q=\frac{2}{3}\end{array}\right.$
Vậy cấp số nhân cần tìm là: $4;6;9$ hoặc $9;6;4$

Bài 2:
Ta có:
     $S_n=nu_1+\frac{(n-1)n}{2}d$
$\Rightarrow 207=3n+\frac{5(n-1)n}{2}$
$\Leftrightarrow 5n^2+n-414=0$
$\Leftrightarrow n=9$, vì $n\in\mathbb{N}$.
Từ đó suy ra: $u_n=u_9=u_1+8d=43$