Use the squeeze theorem to prove the following important trigonometric limit
limθ→0sin(θ)θ=1\lim_{\theta\rightarrow 0} \frac{\sin(\theta)}{\theta} = 1limθ→0θsin(θ)=1
limx→9x−3x−9\lim_{x\rightarrow 9} \frac{\sqrt{x} - 3}{x - 9} limx→9x−9x−3
limx→−3x2−x+12x+3\lim_{x\rightarrow -3} \frac{x^2 - x + 12}{x + 3}limx→−3x+3x2−x+12
limθ→0cos(θ)−1θ\lim_{ \theta\rightarrow 0} \frac{\cos(\theta) - 1}{\theta}limθ→0θcos(θ)−1
Let g(x)=∣x2+x−6∣x−2g(x) = \frac{|x^2 + x - 6|}{x - 2} g(x)=x−2∣x2+x−6∣
Find the limit as x→2+x→2−x→2x\rightarrow 2^{+} x\rightarrow 2^{-} x\rightarrow 2 x→2+x→2−x→2