\(\lim_{x \to 3} (2x + 5)\)
\(\lim_{x \rightarrow 4} \frac{x^2 - 16}{x - 4}\)
Let \( f(x) = \frac{3}{x - 5}\), Evaluate the limit as \(x \rightarrow 5^{-}\) and \( x \rightarrow 5^{+}\)
\(\lim_{x \rightarrow 9} \frac{\sqrt{x} - 3}{x - 9} \)
Let \(g(x) = \frac{|x^2 + x - 6|}{x - 2} \) Find the limit as \(x \rightarrow 2^{+} x \rightarrow 2^{-} x \rightarrow 2 \)
\(\lim_{x \rightarrow -3} \frac{x^2 - x + 12}{x + 3}\)
\(\lim_{t \rightarrow 1} \frac{t^3 - t}{t^2 - 1} \)
\(\lim_{h \rightarrow 0} \frac{(h-5)^2 - 25}{h} \)
\(\lim_{x \rightarrow 0} \frac{\sqrt{2 - x} - \sqrt{2}}{x}\)
Use the squeeze theorem to prove the following important trigonometric limit \(\lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} = 1\)
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