If the function f (x) is differentiable at point XO, then f (x) is continuous at point XO Prove the above proposition

If the function f (x) is differentiable at point XO, then f (x) is continuous at point XO Prove the above proposition

Lim △ Y / △ x = f '(x), then △ y is the infinitesimal of the same order of △ x, and y has
Δ y = f '(x) △ x + O (1) △ x O (1) is an infinitesimal, taking the limit on both sides
lim[f(x)-f(x0)]=lim△y=lim[f'(x)△x+o(1)△x]=alim△x+limo(△x)=0
Then f (x) must be continuous

A problem about limit of higher numbers: to find the limit of higher numbers with power exponent by taking logarithm
LIM (x tends to 0) [(a Λ x + B Λ x + C Λ x) / 3] Λ (1 / x) =? The mobile phone is so hard to call. Who can give us some advice?

Lim [(1 + x) ^ (1 / x)] = e (x - > 0) is transformed into Lim [1 + (a ^ x + B ^ x + C ^ C - 3) / 3] ^ (1 / 3) = Lim e ^ [(1 / x) LN (1 + (a ^ x + B ^ x + C ^ C - 3) / 3] = Lim e ^ [(a ^ x + B ^ x + C ^ C

Some problems in the derivation of logarithm of higher numbers
Y = root ((x-1) (X-2) / (x-3)). How is this logarithmic derivative LNY = 1 / 2 [ln (x-1) + ln (X-2) - ln (x-3)]? How do you get that root? How do you get all of them?

ln (x^a) = a lnx ,a=1/2
ln (x/y) = lnx - lny
Take the natural logarithm of the left and right sides at the same time
lny=1/2[ln(x-1)+ln(x-2)-ln(x-3)]