Limits

Finite Approach

\begin{gathered} \lim_{x \to 2} \frac{x^3+2x^2-1}{5-3x} \\ \frac{(2)^3+2(2)^2-1}{5-3(2)} \\ \frac{15}{-1} \\ -15 \end{gathered}

\begin{align} \lim_{x \to -\infty} e^x &= 0 \\ \lim_{x \to \infty} r^x &= 0 \iff r \in (0, 1) \end{align}

\begin{align} n &\in \mathbb{Z} \\ \sin \left( 2 \pi n + \frac{\pi}{2} \right) &= 1 \\ \sin \left( 2 \pi n - \frac{\pi}{2} \right) &= -1 \\ \sin \left( \pi n \right) &= 0 \end{align}

\begin{gathered} \lim_{x \to \infty} \frac{3x^2-2x+3}{x^2+4x+4} \\ \lim_{x \to \infty} \frac{3x^0-2x^{-1}+3x^{-2}}{x^0+4x^{-1}+4x^{-2}} \\ \frac{3}{1} \\ 3 \end{gathered}

\begin{gather} \lim_{x \to \infty} \sqrt{x^2+1} - x \\ \lim_{x \to \infty} \left( \sqrt{x^2+1} - x \right) \left( \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1} + x} \right) \\ \lim_{x \to \infty} \frac{1}{\sqrt{x^2+1} + x} \\ 0 \end{gather}

\begin{gathered} \lim_{x \to 0} x \sin x^{-1} \\ -1 \le \sin x^{-1} \le 1 \\ -|x| \le x \sin x^{-1} \le |x| \\ 0 \le x \sin x^{-1} \le 0 \end{gathered}

\lim_{x \to a} f(x) = f(a)

\begin{gathered} \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \\ \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} \\ \lim_{x \to 2} x + 2 \\ 4 \end{gathered}

Recall that $\lim_{x \to 0} x^2 \sin x^{-1} = 0$ .

\begin{gathered} g(x) = \begin{cases} x^2 \sin x^{-1} & x \neq 0 \\ 0 & x = 0 \end{cases} \end{gathered}

The function is continuous for all $\mathbb{R}$ .

Continuity can be inherited through composition if the outer function is continuous for the range of the inner function.

\begin{gathered} \lim_{x \to \infty} \sin(e^{-x}) \\ \sin \left( \lim_{x \to \infty} e^{-x} \right) \\ \sin 0 \\ 0 \end{gathered}

\begin{gathered} \lim_{\Delta{x} \to 0} \frac{f(x + \Delta{x}) - f(x)}{\Delta{x}} \\ \frac{\Delta{f(x)}}{\Delta{x}} \\ \frac{df}{dx} \end{gathered}

The function $f(x) = |x|$ is not differentiable at $x=0$ as there are different tangent lines from left and right.

Continuity does not imply differentiability but differentiability implies continuity. Differentiable functions are a subset of continuous functions.

If:

\lim_{ x \to a } \frac{f(x)}{g(x)}

approaches $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and:

\lim_{ x \to a } g'(x) \neq 0

then:

\lim_{ x \to a } \frac{f(x)}{g(x)} =\lim_{ x \to a } \frac{f'(x)}{g'(x)}

There exists a limit for $\frac{sin x}{x}$ even though the denominator approaches $0$ .

\begin{gathered} \lim_{ x \to 0 } \frac{\sin x}{x} \\ \lim_{ x \to 0 } \frac{\cos x}{1} \end{gathered}

A similar approach can be applied to logarithms.

\begin{gathered} \lim_{ x \to \infty } \left( x^{-\frac{1}{3}} \log x \right) \\ \lim_{ x \to \infty } -\frac{1}{3} \left( x^{-\frac{4}{3}} \right) (x^{-1}) \end{gathered}

The sandwich theorem can be used to evaluate $\frac{\sin x}{x}$ .

\begin{gathered} \lim_{ x \to \infty } \frac{x +\sin x}{x} \\ \lim_{ x \to \infty } \frac{x}{x} + \frac{\sin x}{x} \\ 1 + \lim_{ x \to \infty } x^{-1} \sin x \\ 1 + 0 [-1, 1] \\ 1 + 0 \end{gathered}