HKDSE Mathematics Question Review （20201107）

Show that when x ->0:

(e^x-1)/x = 1

and

(a^x-1)/x = ln (a)

Solution:

Use the fact that:

e^x = 1/0! + x/1! + x^2/2! + ...

a^x = ln(a) e^x

e^0 = 1 = a^0

And the definition of derivative:

e^0 = lim [h->0] (e^h - e^0)/(h-0)

LHS=RHS

because when h->0 we have

((1+h+h^2/2 + ...) -1)/h

= 1 + h/2! + h^2/3! + ... = 1

note that when x=1

a^x can be rewritten as e^(x lna) to get 1

sorry should be x=0

sorry a^x = ln(a) e^x 係錯的， 我再想想後半部 proof.

e^xlna = 1 + xlna +(xlna)^2/2 + ...

根據前半部 proof 的 logic,

我地會抽取到 lna 呢個 common factor

QED

details:

a^0

= lim [h->0] (e^hlna - e^0lna )/ (hlna -1)

(e^hlna -1)/hlna

= (hlna + (hlna)^2/2! + ...)/hlna

= lna (h/lna + h^2lna/2! + h^3lna^2/3! + ...)

= lna

where a >1

代錯數

應該係呢個limit 才對:

lim [h->0] (f(h) - f(0))/(h-0)

(hlna + (hlna)^2/2! + ...)/h

= lna (1 + hlna/2! + h^2lna^2/3! + ...

= lna (1 + 0 + 0 + ...)

= lna

成日犯 careless manipulation errors

我終於明點解我2007 Pure Math 得 E

我連中五數學都唔記得點做，NOT TO MENTION 中七PUREMATHS

我係有聽書但唔做練習

pure math啲theory我睇

其他同學做mock我就係度煲SCMP

煲極UE都係得個E

theory 我識*