limh→0(h−5)2−25h\lim_{h\rightarrow 0} \frac{(h-5)^2 - 25}{h} limh→0h(h−5)2−25
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
limx→02−x−2x\lim_{x\rightarrow 0} \frac{\sqrt{2 - x} - \sqrt{2}}{x}limx→0x2−x−2
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