We are asked to evaluate the limits: (1) $ \lim_{x \to 0} (\frac{1}{x} \cdot \sin x) $ (2) $ \lim_{x \to 0} (\frac{1}{\tan x} - \frac{1}{x}) $

1. Problem Description

We are asked to evaluate the limits:

(1) $\lim_{x \to 0} (\frac{1}{x} \cdot \sin x)$

(2) $\lim_{x \to 0} (\frac{1}{\tan x} - \frac{1}{x})$

2. Solution Steps

(1) For the first limit, we have:

$\lim_{x \to 0} (\frac{1}{x} \cdot \sin x) = \lim_{x \to 0} \frac{\sin x}{x}$

We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$

(2) For the second limit, we have:

$\lim_{x \to 0} (\frac{1}{\tan x} - \frac{1}{x}) = \lim_{x \to 0} (\frac{x - \tan x}{x \tan x})$

Since $\lim_{x \to 0} (x - \tan x) = 0$ and $\lim_{x \to 0} x \tan x = 0$, we can use L'Hopital's rule.

$\frac{d}{dx} (x - \tan x) = 1 - \sec^2 x = 1 - \frac{1}{\cos^2 x} = \frac{\cos^2 x - 1}{\cos^2 x} = \frac{-\sin^2 x}{\cos^2 x} = -\tan^2 x$

$\frac{d}{dx} (x \tan x) = \tan x + x \sec^2 x$

Therefore,

$\lim_{x \to 0} \frac{x - \tan x}{x \tan x} = \lim_{x \to 0} \frac{-\tan^2 x}{\tan x + x \sec^2 x} = \lim_{x \to 0} \frac{-\tan^2 x}{\tan x + x \sec^2 x}$

Since $\lim_{x \to 0} -\tan^2 x = 0$ and $\lim_{x \to 0} \tan x + x \sec^2 x = 0$, we can apply L'Hopital's rule again.

$\frac{d}{dx} (-\tan^2 x) = -2 \tan x \sec^2 x$

$\frac{d}{dx} (\tan x + x \sec^2 x) = \sec^2 x + \sec^2 x + x (2 \sec x) (\sec x \tan x) = 2 \sec^2 x + 2x \sec^2 x \tan x$

So,

$\lim_{x \to 0} \frac{-\tan^2 x}{\tan x + x \sec^2 x} = \lim_{x \to 0} \frac{-2 \tan x \sec^2 x}{2 \sec^2 x + 2x \sec^2 x \tan x} = \lim_{x \to 0} \frac{- \tan x \sec^2 x}{ \sec^2 x + x \sec^2 x \tan x} = \lim_{x \to 0} \frac{- \tan x}{1 + x \tan x}$

Since $\lim_{x \to 0} - \tan x = 0$ and $\lim_{x \to 0} 1 + x \tan x = 1$, we have:

$\lim_{x \to 0} \frac{- \tan x}{1 + x \tan x} = \frac{0}{1} = 0$

Alternatively, we can use Taylor series expansions:

$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + ...$

So,

$\frac{1}{\tan x} - \frac{1}{x} = \frac{x - \tan x}{x \tan x} = \frac{x - (x + \frac{x^3}{3} + \frac{2x^5}{15} + ...)}{x(x + \frac{x^3}{3} + \frac{2x^5}{15} + ...)} = \frac{-\frac{x^3}{3} - \frac{2x^5}{15} - ...}{x^2 + \frac{x^4}{3} + \frac{2x^6}{15} + ...} = \frac{x^3(-\frac{1}{3} - \frac{2x^2}{15} - ...)}{x^2(1 + \frac{x^2}{3} + \frac{2x^4}{15} + ...)} = x \frac{-\frac{1}{3} - \frac{2x^2}{15} - ...}{1 + \frac{x^2}{3} + \frac{2x^4}{15} + ...}$

As $x \to 0$, this approaches $0 \cdot (-\frac{1}{3}) = 0$.

3. Final Answer

(1) The limit of $\lim_{x \to 0} (\frac{1}{x} \cdot \sin x)$ is

1. (2) The limit of $ \lim_{x \to 0} (\frac{1}{\tan x} - \frac{1}{x}) $ is

0.