 # Maths – Differentiation (1)

##### $$f(x) = \frac{x^{2}-1}{x-1}$$ is defined everywhere except x=1

$$f(1)$$ ?

$$f$$ is not defined at $$x=1$$

When x close to 1, $$x$$ -> $$1$$

f(x) close to 2, >>> $$f(x)$$ -> $$2$$

$$\neq f(1) = 2$$

# Definition of Limit

## Left Limit

$$lim_{x \to c^{-}} f(x) = M$$

Example:

$$lim_{x \to 1^{-}} f(x) = 2$$

## Right Limit

$$lim_{x \to c^{+}} f(x) = N$$

Example: $$lim_{x \to 1^{+}} f(x) =2$$

Limit exists! When M=N

# Existence of the Limit of a Function

$$f(1)=2$$

Left limit: $$lim_{x \to 1^{-}} f(x) = 2$$

Right limit: $$lim_{x \to 1^{+}} f(x) = 2$$

So, $$lim_{x \to 1} f(x) = 2$$

$$g(1)=3$$

Left limit: $$lim_{x \to 1^{-}} g(x) = 2$$

Right limit: $$lim_{x \to 1^{+}} g(x) = 2$$

So, $$lim_{x \to 1} g(x) = 2$$

$$y=f(x)$$

$$f(-2) = 2$$ , just look at the black dot, there is no explanation for why “2”, this is how it works

Left limit: $$lim_{x \to -2^{-}} f(x) = 1$$

Right limit: $$lim_{x \to -2^{+}} f(x) = 3$$

$$lim_{x \to -2} f(x) =$$does not exist as left $$lim \neq$$ right $$lim$$.

ONLY left limit equals to the right limit, limit exist.

Left limit when $$x \to 0^{-}$$

Keep trying values from the left.

$$\to lim_{x \to 0^{-}} f(x) = -\infty$$

Right limit when $$x \to 0^{+}$$

$$\to lim_{x \to 0^{+}} f(x) = +\infty$$

$$-\infty \neq +\infty$$, therefore, limit does not exist.

Left limit when $$x \to 1^{-}$$

$$\to lim_{x \to 1^{-}} f(x) = 2$$

Right limit when $$x \to 1^{+}$$

$$\to lim_{x \to 1^{+}} f(x) = 2$$

$$lim_{x \to 1} f(x) =2$$

### We will not always check left lim and right lim in later, therefore, we have some tricks to check them.

$$lim_{x \to 1} \frac{(3x+2)(x-2)}{5x-2} = f(x)$$

not defined when $$x = \frac{2}{5}$$

Find limit:
$$\frac{[3(1)+2][1-2]}{5(1)-2}$$

$$= -\frac{5}{3}$$

Cannot directly sub 3 into x to check as it will become 0.

$$lim_{x \to 3} \frac{x^{2}-2x-3}{x-3}$$

$$= lim_{x \to 3} \frac{(x-3)(x+1)}{x-3}$$

$$= (3+1)$$

$$= 4$$

#### It is nearly impossible to have $$\frac{0}{0}$$ in any homework, quiz and exam, this is basically inviting you to factorise.

$$lim_{x \to 0} \frac{\sqrt{x+4}-2}{x}$$

$$= lim_{x \to 0} \frac{(\sqrt{x+4}-2)}{x} x \frac{(\sqrt{x+4}+2)}{\sqrt{x+4}+2}$$

$$= lim_{x \to 0} \frac{(x+4)-4}{x(\sqrt{x+4}+2)}$$

$$= lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)}$$

$$= lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)}$$

$$= lim_{x \to 0} \frac{1}{(\sqrt{x+4}+2)}$$

$$= \frac{1}{(\sqrt{0+4}+2)} = \frac{1}{4}$$

$$lim_{x \to \infty} \frac{1}{x}$$ :it will get smaller and smaller, and 0 finally.

$$lim_{x \to \infty} \frac{1}{x} = 0$$

$$f(x) = \frac{1}{x}$$

$$f(100) = \frac{1}{100}$$

$$f(1000) = \frac{1}{1000}$$

$$g(x)=x$$

$$lim_{x \to \infty} g(x) = +\infty$$

$$lim_{x \to +\infty} [\frac{1}{2^{x}} + \frac{1}{x^{2}+x}]$$

$$lim_{x \to +\infty}[\frac{1}{2^{x}}] + lim_{x \to +\infty}[\frac{1}{x^{2}+x}]$$

$$0+0=0$$

Example:

$$f(10) = \frac{1}{2^{10}} + \frac{1}{10^{2} + 10}$$

$$lim_{x \to \infty} [x^{4} – \pi x^{2}]$$

$$= lim_{x \to \infty} x^{4} – lim_{x \to \infty} \pi x^{2}$$

### WRONG!!!

$$~ lim_{x \to \infty} x^{4} = \infty$$

$$\to$$ limit does not exist.

$$lim_{x \to \infty} \frac{2x^{3}+3}{3x^{3}+1}$$

Tips: always the guy with the highest power.

$$= lim_{x \to \infty} (\frac{2x^{3}+x}{3x^{3}+1}) (\frac{\frac{1}{x^{3}}}{\frac{1}{x^{3}}})$$

— As of 2023-09-19 19:30

___ Start from 2023-09-26 16:50:23

Slope of any function = derivative: $$\frac{df}{dx} = f(x) = y’$$

Differentiation Rules:

1. $$\frac{d}{dx}(C) = 0$$
2. $$\frac{d}{dx}(x) = 1$$
3. $$\frac{d}{dx}(x^{n})=nx^{n-1}$$
4. $$\frac{d}{dx}[u(x)+v(x)] = \frac{d}{dx}u(x) + \frac{d}{dx}v(x)$$
5. $$\frac{d}{dx}[Cu(x)] = C \frac{d}{dx}u(x)$$, C is constant.