Limit proof Lim1 / xsin1 / x = 0 (x approaches infinity)

Limit proof Lim1 / xsin1 / x = 0 (x approaches infinity)

because
Lim1 / x = 0 (x approaches infinity)
and
Sin1 / X is a bounded function
therefore
Original function limit = 0

Let f (x) = 2 ^ (1 / x-1). It is proved that the limit of F (x) does not exist when x → 1

The left limit is LIM (x → 1 -) 2 ^ (1 / (x-1)) (= 2 ^ (- ∞)) = 0
The right limit is LIM (x → 1 +) 2 ^ (1 / (x-1)) (= 2 ^ (+ ∞)) = + ∞
So there is no limit

High number limit: X -- > infinity limf (x) = (1 + 1 / x) ^ x = e. it seems that it cannot be proved by exponentially logarithmicizing f (x). Which step is the problem

This is the standard form of the infinite power of 1. You can rewrite (1 + 1 / x) ^ x into XLN (1 + 1 / x), and ln (1 + 1 / x) is equivalent to 1 / X when X - > infinity. This is equivalent to infinitesimal substitution. In this way, XLN (1 + 1 / x) becomes x * 1 / x = 1, so X -- > infinity limf (x) = (1 + 1 / x) ^ x = e baoji0725 children's shoes

Let f (0) = 0 and the limit x → 0, IMF (x) / X exist, then x → 0, limf (x) / X=

The result should be 1

Find the limit LIM (e ^ x-e ^-x) / X Please help! What's the process?

There may be the following two cases: 1. X - > ∞, at this time, the numerator / denominator is of ∞ / ∞ type. According to lobida's law, the derivative of numerator and denominator can be obtained at the same time, Lim X - > ∞ (e ^ x + e ^-x) = ∞; 2. X - > 0. At this time, the numerator / denominator is of type 0 / 0. According to lobida's law, the derivation of numerator and denominator can be obtained at the same time. Lim X - > 0 (e ^ x + e ^ - x) = 2. I don't know the upper

LIM (x → 0) e ^ (- 1 / x ^ 2) limit?

e^(-1/x^2)=1/[e^(1/x^2)]
If x → 0, then x ^ 2 → 0,1 / x ^ 2 →∞
Because E > 0, e ^ (1 / x ^ 2) →∞
Then 1 / [e ^ (1 / x ^ 2)] → 0
So LIM (x → 0) e ^ (- 1 / x ^ 2) = 0