Finding the limit process of (sin2x-1) / X when the limit x tends to 0 Find the limit of (sin2x-1) / X when x tends to 0 process

When x → 0, sin (2x) and 2x are infinitesimals,
The original formula = LIM (x → 0) ((2x-1) / x) = ∞
The limit of the original formula does not exist

How to find the limit of X * sin2x / (x ^ 2 + 1) when x tends to infinity

x→∞
lim x*sin2x / (x^2+1)
because
2 x sin is bounded
X / (x ^ 2 + 1) = 1 / (x + (1 / x)) tends to zero and is infinitesimal
Infinitesimal multiplied by bounded quantity is infinitesimal
Therefore,
lim x*sin2x / (x^2+1)=0
If you don't understand, please ask

The limit of sin2x / 2cos (π - x) when x tends to π / 4

sin2x/2cos(π-x)=2sinxcosx/(-2cosx)=-sinx
limx→π/4[sin2x/2cos(π-x)]=limx→π/4)[-sinx]=-sinπ/4=-√2/2

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