Limx → 0 {√ (x + 1) - 1} / sin2x

Limx → 0 {√ (x + 1) - 1} / sin2x

limx→0{√(x+1)-1}/sin2x=limx→0（x/sin2x)*1/({√(x+1)+1)=2/2=1

limx→0+ln(1+sin2x)/x
What is ln (1 + sin2x) / X equal to when x right approaches 0

It's type 0 / 0, using the lobita rule:
lim ln(1+sin2x)/x
x->0+
=lim 1/(1+sin2x)*cos2x*2 /1
x->0+
= 1/(1+0)*1*2/1=1/2

limx=√（x+1）- √（1+x²） x--0 —————————— √（1+x）- 1
Dizzy baidu this typesetting ah
Sorry to write it again
lim（x→0） =【√（x+1）- √（1+x²）】/【√（1+x）- 1】

I remind you, you don't ask
The original formula = Lim [(x + 1) - (1 + X & # 178;)] / [√ (1 + x) - 1] [√ (x + 1) + √ (1 + X & # 178;)]
=lim(x-x^2)/[√（1+x）- 1][√（x+1）+ √（1+x²）]
When X - > 0
Lim [√ (1 + x) - 1] = limx / 2 = 0 (that is, [√ (1 + x) - 1] and X / 2 are equivalent infinitesimals, which is the knowledge to be known)
The theorem of Equivalent Infinitesimal Substitution
The original formula = limx (1-x) / {X / 2 [√ (x + 1) + √ (1 + X & # 178;)]}
=lim2(1-x)/[√（x+1）+ √（1+x²）]
=lim2*lim(1-x)/[lim√（x+1）+lim√（1+x²）]
=2*1/(√1+√1)
=2/2
=1