Calculus 1

$\int \frac{6}{\sqrt{x-9x^2}}\ dx$

What is the indefinite integral of $y = 100x^2$

$\lim_{x \to 3} (2x + 5)$

$\lim_{x \rightarrow 4} \frac{x^2 - 16}{x - 4}$

$\lim_{x\rightarrow 9} \frac{\sqrt{x} - 3}{x - 9}$

$\lim_{x\rightarrow -3} \frac{x^2 - x + 12}{x + 3}$

$\lim_{t \rightarrow 1} \frac{t^3 - t}{t^2 - 1}$

$\lim_{h\rightarrow 0} \frac{(h-5)^2 - 25}{h}$

$\lim_{x\rightarrow 0} \frac{\sqrt{2 - x} - \sqrt{2}}{x}$

Use the squeeze theorem to prove the following important trigonometric limit

$\lim_{\theta\rightarrow 0} \frac{\sin(\theta)}{\theta} = 1$

$\lim_{ \theta\rightarrow 0} \frac{\cos(\theta) - 1}{\theta}$

Let $g(x) = \frac{|x^2 + x - 6|}{x - 2}$

Find the limit as $x\rightarrow 2^{+} x\rightarrow 2^{-} x\rightarrow 2$

Let $f(x) = \frac{3}{x - 5}$,

Evaluate the limit as $x\rightarrow 5^{-}$ and $x\rightarrow 5^{+}$

$\lim_{x\rightarrow \infty}\frac{2x - 1}{x + 1}$

$\lim_{x\rightarrow \infty}\frac{3x^2 - 5x + 1}{x^3 - 1}$

$\lim_{x\rightarrow \infty} \frac{2x + 1}{\sqrt{x^2 - x}}$

$\lim_{\rightarrow \infty}\sin\frac{1}{x}$

$\lim_{x\rightarrow \infty}(\frac{5}{x} - \arctan{x})$

$\lim_{x\rightarrow -\infty} \frac{x}{\sqrt{x^2 + 1}}$

$\lim_{\theta\rightarrow 0} \frac{\sec{(2\theta)} \tan{(3 \theta)}}{5 \theta}$