Find the limit: x ^ - 1-e of LIM (x-0) (1 + x) divided by X

Find the limit: x ^ - 1-e of LIM (x-0) (1 + x) divided by X

Using Robida's Law: LIM (x - > 0) ((1 + x) ^ (1 / x) - E) / x = LIM (x - > 0) ((1 + x) ^ (1 / x)) 'let y = (1 + x) ^ (1 / X), LNY = ln (x + 1) / x, derive: y' / y = (x / (x + 1) - ln (x + 1)) / x ^ 2 = (x - (x + 1) ln (x + 1)) / (x ^ 2 (1 + x)) limy '= limylim (x - (x + 1) ln (x + 1)) / (x ^ 2 (1 + x)) = e

Find the limit Lim x →∞ X-1 divided by 3x-x-1

Lim x →∞ x-square-1 divided by 3x-square-x-1
=Lim x →∞ 1-1 / x square divided by 3-1 / X-1 / x square
=（1-0）÷（3-0-0）
=1/3

When x = 0, f (x) = 1. When x is not equal to 0, f (x) = the absolute value of SiNx. Find the limit at x = 0? How?

Function is a discontinuous function with jump discontinuities
When x tends to - 0, | SiNx | tends to 0, and when x tends to + 0, | SiNx | also tends to zero, so the limit of X at 0 is zero
The limit is not equal to the value of the function, so it is a discontinuous function

It is proved that f (x) = absolute value x, and the limit is zero when x tends to 0

Because f (x) = x (x > 0), f (x) = - x (x)

X tends to infinity, and the limit of F (x) is equal to a (a > 0). It is proved that there is a certain M. when x > m, the absolute value of F (x) is greater than half a

Because limf = a
So we can always find a real number set D, which holds when x > M
|F-A|M
There are | f | = | a + (F-A) | > = | a | - | F-A > | a | - | a | / 2 = | a | / 2

What is the absolute value of SiNx divided by X? What is the limit that x approaches zero

Let f (x) = SiNx / |x|
Then LIM (x → 0 +) f (x)
=lim(x→0+)sinx/x
=1
lim(x→0-)f(x)
=lim(x→0-)sinx/(-x)
=-1
The left and right limits are not equal
So the limit doesn't exist