Find the change of function value, vertex coordinate and image of y = - 3 (x + 1) ∧ 2 - 4

Find the change of function value, vertex coordinate and image of y = - 3 (x + 1) ∧ 2 - 4

Find the vertex coordinates of the function y = 2 (x-1) ∧ 2 + 3, the change of the function value and the image.

X tends to be 0 positive, find the limit of [(1 + x) ^ (1 / x) / E] ^ (1 / x)
The answer should be e ^ (- 1 / 2),

Lim [(1 + x) ^ (1 / x) / E] ^ (1 / x) = e ^ {limln [(1 + x) ^ (1 / x) / E] / X} = e ^ {LIM ([ln (1 + x)] / x-1) / X} = e ^ {LIM (1 + x) - x) / X & sup2;} lobita's law = e ^ {LIM (1 / (1 + x) - 1) / (2x)} once more = e ^ {Lim [- 1 / (1 + x) & sup2;] / 2} = e ^ ([- 1 / (1 + 0) & sup2;] / 2) = e ^ (- 1

(urgent) limit problem: X tends to be 0 positive, find the limit of [(1 + x) ^ (1 / x) / E] ^ (1 / x)
X tends to be 0 positive, find the limit of [(1 + x) ^ (1 / x) / E] ^ (1 / x)
Why can't we use (1 + x) ^ (1 / x) = e to get LIM (E / E) ^ 1 / x
If the problem is solved, you can still get + points
I just don't know why we can't directly use LIM (1 + x) ^ (1 / x) = e to bring in the original formula, but first convert the formula into {1 + [(1 + x) ^ (1 / x) - e] / E} {E / [(1 + x) ^ 1 / x] - e} * {[(1 + x) ^ 1 / x] - e} / ex, and then take 1 + [(1 + x) ^ (1 / x) - e] / E as t, and use LIM (1 + T) ^ (1 / T) = e to bring in the formula to solve the problem!
(don't tell me about the process of solving the problem. I just want to know why I can't do it.)
As Qin Keqing said, LIM (1 + x) ^ (1 / x) = e is only true under the limit. This is to seek the limit,
According to jinghuawangzi, X must marry the limit at the same time, but in the book, the formula is changed into {1 + [(1 + x) ^ (1 / x) - e] / E} ^ {E / [(1 + x) ^ 1 / x] - e} * {[(1 + x) ^ 1 / x] - e} / ex, then 1 + [(1 + x) ^ (1 / x) - e] / E is taken as t, and the original formula is lime ^ [(1 + x) 1 / x] / ex. he doesn't marry the limit at the same time, but he still leaves a [(1 + x) 1 / x] / ex.
We can't use the equivalent infinitesimal for the power of the quiet and settled, [(1 + x) 1 / x] / ex, isn't this also the power
Don't think I'm wordy. This problem has bothered me for 2 days. I can't figure it out. I'm rather stupid.
Haha, after asking the students, I finally understand what Qin Keqing said. LIM (1 + x) ^ (1 / x) = e is only true under the limit

(1 + x) ^ (1 / x) = e only holds in the limit state,
If you can substitute LIM (1 + x) ^ (1 / x) = (1 + 0) ^ (1 / x) = 1, it is obviously wrong
X tends to be zero positive, Lim [(1 + x) ^ (1 / x) / E] ^ (1 / x)
=
X approaches positive infinity, Lim [(1 + 1 / x) ^ (x) / E] ^ X
=
lim[(1+1/x)^(xx) / e^x
=
lim e ^ (xxln(1+1/x) - x)
=
e ^ (lim(xxln(1+1/x) - x))
The limit of index is based on the law of lobita

When x approaches 0, is there a limit for (1 + 1 / x) ^ x? If so, what is it?

 
The second equal sign uses the variable substitution t = 1 / x, and the third equal sign uses the law of Robida

Find ((a * x + b * x + C * x) / 3) * (1 / x) when x approaches the limit of 0

Logarithm first
ln(a^x+b^x+c^x)/3)^(1/x)
=[ln((a^x+b^x+c^x)/3)]/x
Lobita, simultaneous derivation up and down:
[3/(a^x+b^x+c^x)]*1/3*(a^xlna+b^xlnb+c^xlnc)
X tends to zero
a^x→1 b^x→1 c^x→1
Then the above formula = 1 / 3 (LNA + LNB + LNC) = ln (ABC) ^ (1 / 3)
Then take e ^ ln (ABC) ^ (1 / 3) = (ABC) ^ 1 / 3
So the limit is (ABC) ^ 1 / 3

Finding the limit Lim x tends to infinity x ^ 2 (1-xsin1 / x)

Find the limit x →∞ Lim {X & # 178;} such that 1 / x = u, then x = 1 / u, when x →∞ u → 0. So the original formula = u → 0lim {[1 - (1 / U) sinu] / U & # 178;} = u → 0lim [(u-sinu) / U & # 179;] = u → 0lim [(1-cosu) / (3U & # 178;)] = u → 0lim [(sinu) / 6U] = u → 0lim [(COSU) / 6] = 1 / 6

Is there a limit when LNX approaches zero?
Such as the title
If so, what is the limit?
Can't it be written as negative infinity?
Isn't it true that there is no limit only when the function is undefined?

No, it is obvious that if you draw a graph, Y axis is an asymptote of y = LNX image, when x approaches 0, the corresponding function value approaches infinitesimal

(x + LNX) ^ 1 / (1-x ^ 2) find the limit when x approaches 1

This kind of problem is very simple. We assume that y = (x + LNX) ^ 1 / (1-x ^ 2) and then find the limit of e ^ LNY. Obviously, LNY = (1 / (1-x ^ 2)) ln (x + LNX) = ln (x + LNX) / (1-x ^ 2) the denominators of molecules tend to 0, so we can get the limit that the molecules are equal to (1 + 1 / x) / (x + LNX) 2 and the denominators are equal to - 2x ~ - 2 by using the law of Robida

Find the limit of (lnx-1) / (x-e) when x approaches E

This problem is infinitesimal / infinitesimal infinitive, you can use two methods
1. The equivalent infinitesimal substitution method is used;
2. Apply the law of derivation
The specific answers are as follows:

Finding limit LIM (x → 1) (x-1) ^ 3 / (LNX) ^ 3=

According to the law of Robida: LIM (x → 1) (x-1) / (LNX) = LIM (x → 1) 1 / (1 / x) = 1
∴lim (x→1) (x-1)^3/(lnx)^3= [lim (x→1) (x-1)/(lnx)]³=1³=1