Solve this: Q (9) limθ→08 sinθ-θcosθ3tanθ+θ2 - Maths -

Question

Solve this:
Q (9) lim θ → 0 8   sin θ - θ cos θ
3 tan θ + θ 2

Neha Sethi · Accepted Answer

Dear student
limθ→08sinθ-θ cosθ3tanθ+θ2=limθ→08sinθtanθ-θ cosθtanθ3tanθtanθ+θ2tanθ=limθ→08sinθtanθ-θ cosθtanθ3+θ2tanθ=limθ→0 8sinθtanθ-limθ→0 θ cosθtanθlimθ→0 3+θ2tanθConsider,limθ→0 8sinθtanθ=8sin(0)tan(0)=8and, limθ→0 θ cosθtanθ=limθ→0cosθ×limθ→0 θtanθ=cos(0) ×limθ→0ddθ(θ)ddθtanθ   By applying L'Hopital's rule in limθ→0 θtanθ=1×limθ→01sec2θ=1sec20=1Consider, limθ→03+θ2tanθ=3So,limθ→0 8sinθtanθ-limθ→0 θ cosθtanθlimθ→0 3+θ2tanθ=8-13=73
Regards

Solve this: Q.(9) lim θ → 0 8 sin θ - θ cos θ 3 tan θ + θ 2

Solve this:
Q.(9) $\lim_{θ \to 0} \frac{8 \sin θ - θ \cos θ}{3 \tan θ + θ^{2}}$