If LIM (x) tends to 0 and f (2x) / x = 1, and f (x) is continuous, then f '(0)=

If LIM (x) tends to 0 and f (2x) / x = 1, and f (x) is continuous, then f '(0)=

LIM (x) tends to 0, f (2x) / x = 1, f (x) is continuous, then f (0) = 0
f'(0)=lim [f(2x)-f(0)]/[2x-0]=lim f(2x)/(2x)=1/2

lim（x→0）（e∧x+x）∧1/x

When x → 0, LIM (e ^ x + x) ^ (1 / x) = lime ^ ln (e ^ x + x) ^ (1 / x) = e ^ limln (e ^ x + x) ^ (1 / x), and limln (e ^ x + x) ^ (1 / x) = Lim [ln (e ^ x + x)] / X,
Using the lobita rule Lim [ln (e ^ x + x)] / x = Lim [ln (e ^ x + x)] '/ X' = LIM (e ^ x + 1) / (e ^ x + x) = (1 + 1) / (1 + 0) = 2,
So the original formula = e ^ 2

LIM (x tends to 0) (x ^ 2 * e ^ (1 / x ^ 2))

LIM (x tends to 0) x ^ 2 * e ^ (1 / x ^ 2)
=When LIM (x tends to 0) e ^ (1 / x ^ 2) / (1 / x ^ 2) x tends to 0, the numerator denominator tends to 0, and the derivative can be obtained by using the law of lobita
=LIM (x tends to 0) e ^ (1 / x ^ 2) * (1 / x ^ 2) '/ (1 / x ^ 2)'
=LIM (x tends to 0) e ^ (1 / x ^ 2)
So obviously, when x tends to zero, 1 / x ^ 2 tends to be positive infinity, so e ^ (1 / x ^ 2) tends to be positive infinity,
That is, the original limit tends to be positive infinity