Limx → 0 ln (2-x ^ 2) - LN2) / cosx-1 limit

Limx → 0 ln (2-x ^ 2) - LN2) / cosx-1 limit

limx→0 ln(2-x^2)-ln2)/cosx-1
=lim(x→0) ln(2-x^2)-ln2)/(x^2/2)
=lim(x→0) ln[(2-x^2)/2]/(x^2/2)
=lim(x→0) ln(1-x^2/2)/(x^2/2)
=lim(x→0) (-x^2/2)/(x^2/2)
=-1

Find the limit limx approaching 0 (1-cosx) / (1-e ^ x)

X → 0, Cox → 1, e ^ x → 1, so the numerator and denominator are close to 0
So you can use lobida's law
Derivative the numerator and denominator separately
Original limit = limx → 0 (SiNx / - e ^ x) = 0 / - 1 = 0

Find the limit LIM (x ~ 0) ((e ^ x + e ^ 2x + e ^ 3x) / 3) ^ 1 / X s

lim(x~0)((e^x+e^2x+e^3x)/3)^1/x
=lim(x~0)(e^(ln(e^x+e^2x+e^3x)/3)/x)
=e^(lim(x~0)(ln(e^x+e^2x+e^3x)/3)/x)
=e^(lim(x~0)(3(e^x+2e^2x+3e^3x)/(e^x+e^2x+e^3x))
=e^(3(1+2+3)/(1+1+1))
=e^6

If f (x) simply subtracts on [a, infinity), the integral on [a, infinity): (integral sign) f (x) DX converges, it is proved that LIM XF (x) = 0 when x tends to infinity;

According to the necessary and sufficient conditions of generalized integral, the integral f (x) DX converges Lim XLN (s) f (x) = m (bounded)
So Lim XF (x) = 0

On the definition of limit distribution and corresponding application examples, I hope you can give advice!

If the random variable {xn} converges to a variable x when n approaches infinity, the distribution of X is called limit distribution
For example, when the sample size approaches infinity, its limit distribution is normal distribution

EN = (- 1 / 3) ^ n is there a limit

because
q=-1/3
and
|q|<1
therefore
Limit = 0