lim┬(x→π)〖(π-x)cos 1/(x-π)〗

lim┬(x→π)〖(π-x)cos 1/(x-π)〗

lim┬(x→π)〖(π-x)cos 1/(x-π)〗=-lim┬(x→π)cos 1=-cos 1

10、lim (x→+∞) cos (1-x/1+x)=

Divide up and down by X
Original formula = LIM (x → + ∞) cos [(1 / x-1) / (1 / x + 1)]
=cos(-1)
=cos1

Find the limit of LIM (x - > 0)?

lim(x→0) [cos(3x)－cos(5x)] / x^2
＝lim(x→0) [2 × sin(4x) × sinx] / x^2
＝lim(x→0) [2 × 4x × x] / x^2
＝8
Using: when x → 0, SiNx and X are equivalent infinitesimals, so sin (4x) is equivalent to 4x

Lim [(3x) ^ 1 / 2] * e ^ (COS (8pi / x) (X -- > + 0) find the limit?

∵ cos (8 π / x) ≤ 1 (x ≠ 0), e ^ x increases monotonically;
∴ 0 < e^(cos(8π/x)) ≤ e^1 = e
∴ 0 < [(3x)^1/2] * e^(cos(8π/x) ≤ [(3x)^1/2]
∵ LIM (x - > 0 +) [(3x) ^ 1 / 2] = 0, by the pinch theorem:
∴ lim(x->0+) [(3x)^1/2] * e^(cos(8π/x) = 0

Limit of LIM (x →∞) (x ^ 2 + x) / (x ^ 4-3x-1)

Both numerator and denominator are divided by the fourth power of X
lim (1/x^2+1/x^3 )/(1-3/x^3 -1/x^4)
=0/1
=0

Lim x → 0 √ (1-cos x) / X limit

This limit does not exist. The left limit is not equal to the right limit