LIM (x → 0) (x ^ 2) [e ^ {(1 / x ^ 2)}] using l'obita's law to find the limit

LIM (x → 0) (x ^ 2) [e ^ {(1 / x ^ 2)}] using l'obita's law to find the limit

Let 1 / x ^ 2 = t, then t tends to be positive infinity
lim(x→0)(x^2)[e^{(1/x^2) }]
=LIM (t → positive infinity) e ^ t / T
=LIM (t → positive infinity) e ^ t
=Positive infinity

What is the limit of 1 / x power of e when x tends to be negative infinity I figured it was 1, but it seems to be 0

Let e ^ (1 / x) = y
lny=1/x
When x tends to be negative infinity, the right side is 0, so y = 1, or e ^ (1 / x) = n √ e, that is, e to the nth power, then when n tends to infinity, it is 1

What is the limit of (1 + 2 / x) to the power of X, where x tends to infinity?

Let 1 / a = 2 / X
Then a →∞
x=2a
The original formula = LIM (a →∞) (1 + 1 / a) ^ 2A
=lim(a→∞)[(1+1/a)^a]²
=e²

The limit Lim x tends to the power of 0 (1-1 / x)?

There's no need for the law of lophida
Because there are two important limits
Lim x tends to the power of 0 (1 + 1 / x) = E
And lim x tends to the x power of 0 (1 + 1 / x) * (1-1 / x) = 1
So Lim x tends to 0 (1-1 / x) to the power of x = 1 / E

The 1 / x power of LIM (x + e's x power) = what x tends to 0

0

0

Let a = (1 + x) ^ (1 / x ^ 2) / e ^ (1 / x)
Then Lim ln a = Lim ln (1 + x) / x ^ 2 - 1 / X
= lim [ ln(1+x) -x ] /x^2
=- 1 / 2 (lobida's law)
So Lim a = e ^ (- 1 / 2)

Find LIM (x → 0) (SiNx / x) ^ (cosx / 1-cosx)

0

0

lim(x→0)(x-sinx)/[x(1-cosx)]
=LIM (x → 0) (1-cosx) / [(1-cosx) + xsinx] lopital law
=lim(x→0)sinx/[sinx+sinx+xcosx]
=lim(x→0)sinx/[2sinx+xcosx]
=lim(x→0)1/[2+xcosx/sinx]
=lim(x→0)1/lim(x→0)[2+xcosx/sinx]
=1/[2+1]
=1/3
Additional notes:
lim(x→0)xcosx/sinx
=lim(x→0)[cosx-xsinx]/cosx
=[1-0]/1
=1

Lim [(1-cosx) ^ 1 / 2] / SiNx, X tends to 0, find the limit

Replace with equivalent infinitesimal
The original formula = LIM (x → 0) √ (2Sin ^ 2 (x / 2)) / SiNx
=lim(x→0)√2|sin(x/2)|/sinx
Because the right limit is LIM (x → 0 +) √ 2 * sin (x / 2) / SiNx = LIM (x → 0) √ 2 * (x / 2) / 2 = √ 2 / 2
Similarly, the left limit is - √ 2 / 2
So there is no limit

Find the following function limits: LIM (SiNx ^ 3) / [x (1-cosx)], (x → 0)

lim(sinx^3)/[x(1-cosx)],(x→0)
=lim(sinx^3)（1+cosx）/[x(1-cosx)（1+cosx)],(x→0)
=lim(1+cosx)*lim[sinx/x],(x→0)
=1+1=2