Question

lim x tend to 0 :sins -tans over x cube

Answer 1

Lim(x→ 0) (sinx - tanx)/x³

first we check here which form limit exist.
put x = 0
0/0 hence, here we can apply L- Hospital rule , for solving .

Lim(x→ 0) ( sinx - tanx )/x³
differentiate numerator and denominator independently wrt x

Lim(x→0){ cosx - sec²x}/3x²

we Know,
sec²x = 1 + tan²x
put it here,

Lim(x→0) {cosx - 1 - tan²x}/3x²
Lim(x→0){ -tan²x - (1 - cosx)}/3x²

[ (1 - cosx) = 2sin²x/2 use here ]

Lim(x→ 0) { -tan²x - 2sin²x/2 }/3x²

now , use concept
Lim(f(x) →0) tanf(x)/f(x) = 1
and, Lim(f(x) →0) sinf(x)/f(x) = 1

Lim(x→0) { -(tanx/x)²×x²- 2(sinx/2/x/2)² ×x²/4 }/3x²

Lim(x→0) { -x² - x²/2}/3x²

Lim(x→0){ -3x²/6x²} = -1/2

hence, value of limit = -1/2