Limits

To find the limit, we can factor the numerator and simplify the expression: (f(x) = \frac{(x + 2)(x - 2)}{x - 2} = x + 2). As (x) approaches (2), (f(x)) approaches (2 + 2 = 4).

As (x) becomes very large, the term (x^2) dominates the expression, and the other terms become insignificant. Therefore, the limit is (\lim_{x \to \infty} \sqrt{x^2 + 1} - x = \lim_{x \to \infty} x\left(\sqrt{1 + \frac{1}{x^2}} - 1\right) = \infty).

We can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{3\cos(3x)}{1} = 3).

We can rewrite the expression as (\frac{e^{2x} - 1}{x} = \frac{e^{2x} - e^0}{x - 0} = \frac{e^{2x} - e^0}{x} \cdot \frac{e^0 + e^0}{e^0 + e^0} = \frac{e^{2x} + 1}{x(e^0 + e^0)}). As (x) approaches (0), the numerator approaches (e^0 + 1 = 2), and the denominator approaches (2e^0 = 2). Therefore, the limit is (2).

Using L'Hopital's rule, we find (\lim_{x \to \infty} \frac{\ln(x + 1)}{x} = \lim_{x \to \infty} \frac{\frac{1}{x + 1}}{1} = \lim_{x \to \infty} \frac{1}{x + 1} = 0).

We can factor the numerator as (x^3 - 8 = (x - 2)(x^2 + 2x + 4)). Substituting (x = 2) into the expression, we get (\frac{(2)^3 - 8}{2 - 2} = \frac{8 - 8}{0} = \frac{0}{0}). Since this is an indeterminate form, we can use L'Hopital's rule to find the limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 2} \frac{x^3 - 8}{x - 2} = \lim_{x \to 2} \frac{3x^2}{1} = 12).

We can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 0} \frac{\tan(2x)}{\sin(3x)} = \lim_{x \to 0} \frac{2\sec^2(2x)}{3\cos(3x)} = \frac{2}{3}).

We can rewrite the expression as (\frac{\sqrt{x^2 + 9} - 3}{x} = \frac{\sqrt{x^2 + 9} - 3}{x} \cdot \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} = \frac{(x^2 + 9) - 9}{x(\sqrt{x^2 + 9} + 3)} = \frac{x^2}{x(\sqrt{x^2 + 9} + 3)}). As (x) approaches (\infty), the term (x^2) dominates the expression, and the other terms become insignificant. Therefore, the limit is (\lim_{x \to \infty} \frac{\sqrt{x^2 + 9} - 3}{x} = \lim_{x \to \infty} \frac{x^2}{x(\sqrt{x^2 + 9} + 3)} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 9} + 3} = 0).

We can factor the numerator as (x^2 - 4x + 3 = (x - 3)(x - 1)). Substituting (x = 3) into the expression, we get (\frac{(3)^2 - 4(3) + 3}{3 - 3} = \frac{9 - 12 + 3}{0} = \frac{0}{0}). Since this is an indeterminate form, we can use L'Hopital's rule to find the limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 3} \frac{x^2 - 4x + 3}{x - 3} = \lim_{x \to 3} \frac{2x - 4}{1} = 2).

We can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 0} \frac{\sin^2(x)}{x} = \lim_{x \to 0} \frac{2\sin(x)\cos(x)}{1} = \lim_{x \to 0} 2\sin(x)\cos(x) = 0).

Using L'Hopital's rule, we find (\lim_{x \to \infty} \frac{\log(x + 1)}{x} = \lim_{x \to \infty} \frac{\frac{1}{x + 1}}{1} = \lim_{x \to \infty} \frac{1}{x + 1} = 0).

We can factor the numerator as (x^3 + 2x^2 - 3x + 4 = (x - 1)(x^2 + 3x - 4)). Substituting (x = 2) into the expression, we get (\frac{(2)^3 + 2(2)^2 - 3(2) + 4}{(2)^2 - 1} = \frac{8 + 8 - 6 + 4}{4 - 1} = \frac{14}{3} = 11).

We can rewrite the expression as (\frac{e^{2x} - e^{-2x}}{e^x} = \frac{e^{2x}}{e^x} - \frac{e^{-2x}}{e^x} = e^x - e^{-x}). As (x) approaches (\infty), the term (e^x) dominates the expression, and the other term becomes insignificant. Therefore, the limit is (\lim_{x \to \infty} \frac{e^{2x} - e^{-2x}}{e^x} = \lim_{x \to \infty} (e^x - e^{-x}) = \infty).

We can simplify the expression by rationalizing the numerator: (\frac{\sqrt{x^2 + 4x + 4} - x}{x + 2} = \frac{\sqrt{(x + 2)^2} - x}{x + 2} = \frac{x + 2 - x}{x + 2} = \frac{2}{x + 2}). Substituting (x = -2) into the expression, we get (\frac{2}{-2 + 2} = \frac{2}{0}), which is an indeterminate form. Therefore, we can use L'Hopital's rule to find the limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to -2} \frac{\sqrt{x^2 + 4x + 4} - x}{x + 2} = \lim_{x \to -2} \frac{\frac{1}{2\sqrt{x^2 + 4x + 4}}(2x + 4)}{1} = \lim_{x \to -2} \frac{2x + 4}{2\sqrt{x^2 + 4x + 4}} = 1).

We can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator, we get (\lim_{x \to 0} \frac{\sin(x) - x}{x^3} = \lim_{x \to 0} \frac{\cos(x) - 1}{3x^2} = \lim_{x \to 0} \frac{-\sin(x)}{6x} = 0).