A question about the limit of higher numbers When x tends to 2, the square of y = x tends to 4. Ask how much "naimu Da" is equal to, then when | X-2|

A question about the limit of higher numbers When x tends to 2, the square of y = x tends to 4. Ask how much "naimu Da" is equal to, then when | X-2|

[i]
① Right ε= 0.001>0 ,
To make: | x ^ 2-4 |< ε= 0.001 established,
If | X-2 | < 1, then | x + 2 | = | 4 + (X-2) | ≤ 4 + | X-2|

A problem of finding limits in higher numbers It is proved by definition that LIM (4x + 1) = 9

prove:
①
To any ε> 0
To make: | (4x + 1) - 9| = 4| X-2|

A limit problem of high numbers Ln (x + 1) / x ^ 2-2x when x → 0 That ^ means square The answer is - 1 / 2 I want to know why

L'h rule
Derivative of numerator and denominator, get
Original formula = [1 / (x + 1)] / (2x-2)
If x = 0, we can get
Original formula = - 1 / 2

Help calculate the following limits LIM (1-5 / x) x is at the top right of the bracket because I don't know how to type it x→0

Using the conclusion: when x tends to positive infinity (1 + 1 / x) ^ x = E
When x tends to positive infinity, the original formula can become Lim [(1 + (- 5 / x)] ^ (- X / 5)] ^ (- 5) = e ^ (- 5)

Calculate the following limits lim x^2+2x-2 x->-1 --------- x^2+1 lim X - > infinite (1 + 1 / x) (2-1 / x ^ 2) lim x->-1 (x^2+2x-2)/(x^2+1)

1.lim(x^2+2x -2)/(x^2+1)=
Just put x = - 1 into the above formula
So: LIM (x ^ 2 + 2x - 2) / (x ^ 2 + 1) = (1-2-2) / (1 + 1) = - 3 / 2
2.lim
X - > infinite (1 + 1 / x) (2-1 / x ^ 2)
When x tends to ∞, 1 / X tends to 0.1/x ^ 2 tends to 0
So: Lim
X - > infinite (1 + 1 / x) (2-1 / x ^ 2) = (1 + 0) (2-0) = 2

①limx→0(x+e^3x)^1/x ②limx→0(e^x－e^sinx)/(sinx)^3 ③limx→0(e^-1/x^2)/x^100

①limx→0(x+e^3x)^1/x
=lim[e^ln(x+e^3x)^1/x
=e^lim[ln(x+e^3x)/x]
=E ^ Lim [(1 + 3E ^ 3x) / (x + e ^ 3x)] rubida
=e^4
②limx→0(e^x－e^sinx)/(sinx)^3
=lime^x[1－e^(sinx-x)]/(sinx)^3
=Lime ^ x * Lim [sinx-x] / x ^ 3 equivalent replacement + four arithmetic operations
=Lim [cosx-1] / 3x ^ 2 rubida
=Lim SiNx / 6x Robida
=-1/6
③limx→0(e^-1/x^2)/x^100
=LIM (e ^ - 1 / x ^ 2) / x ^ 100 let 1 / x ^ 2 = t
=lim(t→+∞)(e^-t)*t^50
= lim t^50/e^t
=0 uses 50 times of Robida's law