lim 1/[(sinx)^2-x^2]= x->0 Such as title

lim 1/[(sinx)^2-x^2]= x->0 Such as title

lim( x->0)1/(sin ² x-x ²) → - ∞, the limit does not exist
lim( x->∞)1/(sin ² x-x ²)= 0

Find LIM (x → 0) [√ (1 + TaNx) -√ (1 + SiNx)] / X Ask the great God!, I figured out that the answer was 0. It felt strange

LIM (x → 0) [√ (1 + TaNx) -√ (1 + SiNx)] / X numerator denominator multiplied by [√ (1 + TaNx) + √ (1 + SiNx)] = LIM (x → 0) [√ (1 + TaNx) -√ (1 + SiNx)] * [√ (1 + TaNx) + √ (1 + SiNx)] / [√ (1 + TaNx) + √ (1 + SiNx)] * x = LIM (x → 0) (TaNx - SiNx) / [√ (1 + TaNx) + √

lim (x→0) (e^x-x)^(1/sinx)

Consider using ln to simplify the limit
Original formula = Lim e ^ [ln (e ^ x-x) ^ (1 / SiNx)]
=Lim e ^ [ln (e ^ x-x) / SiNx]
=Lim e ^ [(e ^ x-1) / (e ^ x-x) / cosx] [ln (e ^ x-x) / SiNx uses lobida's Law]
=lim e^0
=1
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Find the limit LIM (x - > 0) (1 / sinx-1 / x)

Lobida's law, (x-sinx) / (xsinx) = (1-cosx) / (SiNx + xcosx) = SiNx / (2cosx xsinx) = 0

It is proved that LIM (x - + ∞) (SiNx) / (√ x) = 0 Please be more detailed. I know it's very simple

SiNx is a bounded function when x → + ∞, so LIM (x → + ∞) (SiNx) / (√ x) = 0

lim(x→0)[e^x-e^(-x)]/sinx=?

Use Robida's law, that is, the derivation of numerator and denominator at the same time! (type 0 / 0)
Original formula = LIM (x → 0) [e ^ x-e ^ (- x)] / SiNx
=LIM (x → 0) [e ^ x + e ^ (- x)] / cosx (lobida's law)
=LIM (x → 0) [e ^ 0 + e ^ (- 0)] / cos0 (bring in x = 0)
=lim(x→0)（1+1）/1
=2