Root 8 - negative root 2 - the zero power of root 2 plus one third of negative first power

Root 8 - negative root 2 - the zero power of root 2 plus one third of negative first power

√8-|-√2|-(√2+2015)^0+(1/3)^(-1)=2√2-√2-1+3=2+√2

What is the quantitative relationship between a, B and C given that the a power of 2 = 3, the B power of 2 = 6, and the C power of 2 = 18

2^a*2^b=2^c
2^(a+b)=2^c
So a + B = C

{[(the third power of (2 root sign 2) + the - 2 power of (- 0.5) - (π - 3.14)} * one tenth of the root sign

The original formula = {[(2 ^ 3 / 2) ^ 1 / 3 + √ 5] + (- 2) ^ 2-1} * √ 10 / 10
={√2+√5+3}*√10/10
=√5/5+√2/2+3√10/10.

The cubic of LIM (x + 1) minus the cubic of (X-2) divided by (the square of x plus 2x plus 3) x → + ∞

The numerator and denominator divide x ^ 3,

The limit of the following function can be obtained by using l'obita's rule. LIM (xcot2x) x → 0

It is changed to X / Tan (2x), i.e. "0 / 0" type, which can be obtained by using the law of robetta
lim_ x/tan(2x)=lim_ 1/(2/cos^2(2x))=lim_ cos^2(2x)/2=1/2

Limx tends to positive infinity to the power of X of (- 1-1 / x)? X tends to infinity. What about the power of lim - (1 + 1 / x) to the power of - X of LIM (1 + 1 / x)? Are these two equal?

lim-(1+1/x)^x=-e
lim(1+1/x)^(-x)=lim1/(1+1/x)^x=1/e
These two limits are different

The limit of LIM (X -- ∞) (sinx-x ^ 2) / (cosx + x ^ 2)

Analysis:
Divide by x 2 at the same time
lim(x→∞)（sinx/x²－1）/（cosx/x²＋1）
Because x →∞, so 1 / x 2 → 0, that is infinitesimal, and SiNx and cosx are bounded functions! So SiNx / x? And cosx / x? Are still infinitesimal!
So the original formula =
lim(x→∞)（0－1）/（0＋1）＝－1.
Methods and skills: if this is a multiple choice question, you can directly remove SiNx and cosx, and it will be
lim(x→∞)(－1)/1＝－1.
This is because when x →∞, other terms run to infinity, only SiNx and cosx are always in the circle between 〔 - 1,1], which can be ignored!

LIM (x + cosx) / (x + SiNx) limit when x tends to infinity What is LIM (1 + (cosx SiNx) / (x + SiNx)), can you explain it?

lim(x+cosx)/(x+sinx)=lim(1+(cosx-sinx)/(x+sinx))
=1

LIM (e ^ (x ^ 2) - 1) / (cosx-1), why is the limit of X → 0 - 2

Replace with equivalent infinitesimal
e^x²-1 ~ x²,cosx-1 ~ -x²/2,x->0
The original limit = limx? 2 / (- x? / 2) = - 2, X - > 0

Find the limit Lim [cosx-e ^ (- x ^ 2 / 2)] / x ^ 4, where x tends to 0 Why am I wrong lim[cosx-e^(-x^2/2)+1-1]/x^4=lim[cosx-1-e^(-x^2/2)+1]]/x^4=lim[-(1-cosx)-(e^(-x^2/2)-1)]]/x^4=lim-(1-cosx)/x^4-lim(e^(-x^2/2)-1)/x^4=lim-1/2x^2/x^4-lim-x^2/x^4=0 That is to add 1 to the denominator and subtract 1, and then use the equivalent infinitesimal,

If there are addition and subtraction, it can't be replaced by equivalent infinitesimal