Evaluating Limits

If \(\lim_{x \to c} f(x)\) and \(\lim_{x \to c} g(x)\) exist, then:
\(\displaystyle \lim_{x \to c}\bigl( kf(x) \bigr) = k \Bigl(\lim_{x \to c} f(x)\Bigr) \,\text{ for } k \in \mathbf{R}\)
\(\displaystyle \lim_{x \to c}\bigl( f(x) \pm g(x) \bigr) = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)\)
\(\displaystyle \lim_{x \to c}\bigl( f(x) \cdot g(x) \bigr) = \Bigl(\lim_{x \to c} f(x)\Bigr) \cdot \Bigl(\lim_{x \to c} g(x)\Bigr)\)
\(\displaystyle \lim_{x \to c}\bigl( f(x) \bigr)^n = \Bigl(\lim_{x \to c} f(x) \Bigr)^n \,\text{ for } n \in \mathbf{R}\)
\(\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to c} f(x)}{\lim\limits_{x \to c} g(x)} \,\text{ so long as } \lim_{x \to c} g(x) \neq 0\)

As exercise, for each of the following limits, either determine its value or decide that the limit does not exist. For any limit whose value, though it exists, proves difficult to express exactly, calculate a decimal approximation to the value accurate to within ±one-thousandth.

\(\displaystyle \lim\limits_{x \to 1^-}\arcsin(x)\)
\(\displaystyle \lim\limits_{x \to 0}\frac{x^4+x^3}{x^4+3x^3}\)
\(\displaystyle \lim\limits_{x \to 3^+}\Bigl(\bigr|x^2+x\bigl|-x\Bigr)\)
\(\displaystyle \lim\limits_{x \to -1^+}\sqrt{1-x^2}\)
\(\displaystyle \lim\limits_{x \to -1^-}\sqrt{1-x^2}\)
\(\displaystyle \lim\limits_{x \to 0^-}\frac{x}{|x|}\)
\(\displaystyle \lim\limits_{x \to 3}\frac{x-3}{x+3}\)
\(\displaystyle \lim\limits_{x \to -3^+}\frac{x-3}{x+3}\)
\(\displaystyle \lim\limits_{x \to -1}\frac{1}{x-1}\)
\(\displaystyle \lim\limits_{x \to 7}\frac{1}{x-7}\)
\(\displaystyle \lim\limits_{x \to 7}\frac{1}{(x-7)^4}\)
\(\displaystyle \lim\limits_{x \to -3}\frac{x^2-9}{x+3}\)
\(\displaystyle \lim\limits_{x \to 3}\frac{x^2-9}{x-3}\)
\(\displaystyle \lim\limits_{x \to -1}\frac{x^2+2x+1}{x+1} \)
\(\displaystyle \lim\limits_{x \to -1}\frac{x^3+1}{x+1} \)
\(\displaystyle \lim\limits_{x \to -1}\frac{x^3+3x-2}{(x+1)^2} \)
\(\displaystyle \lim\limits_{x \to -1}\frac{x^3-6x^2+11x-6}{(x+1)^2} \)
\(\displaystyle \lim\limits_{x \to 4}\frac{x^2+5x-36}{x^2-16}\)
\(\displaystyle \lim\limits_{x \to 25}\frac{x-25}{\sqrt{x}-5}\)
\(\displaystyle \lim\limits_{x \to 3^+}\frac{1}{(x-3)^3}\)
\(\displaystyle \lim\limits_{x \to 3}\frac{\sqrt{x^2+16}}{x-3}\)
\(\displaystyle \lim\limits_{x \to -2}\frac{3x^2-8x-3}{2x^2-18}\)
\(\displaystyle \lim\limits_{x \to 2}\frac{x^2+2x+1}{x^2-2x+1}\)
\(\displaystyle \lim\limits_{x \to 3}\frac{x+3}{x^2-9}\)
\(\displaystyle \lim\limits_{x \to -1}\frac{x+1}{x^2+x}\)
\(\displaystyle \lim\limits_{x \to 1^+}\frac{1}{x^2+1}\)
\(\displaystyle \lim\limits_{x \to 1}\Bigl(x^2+5x-\frac{1}{2-x}\Bigr)\)
\(\displaystyle \lim\limits_{x \to 0}\frac{x^2}{x^2+2x-3}\)
\(\displaystyle \lim\limits_{x \to 1}\frac{x^2-1}{x^2+2x-3}\)
\(\displaystyle \lim\limits_{x \to 1}\frac{5x}{x^2+2x-3}\)
\(\displaystyle \lim\limits_{x \to 7}\frac{\sqrt{x}-\sqrt{7}}{x-7}\)
\(\displaystyle \lim\limits_{x \to 1}\frac{3-\sqrt{8+x}}{x-1}\)
\(\displaystyle \lim\limits_{x \to -2}\frac{x^2+9x+14}{\sqrt{x+11}-3}\)
\(\displaystyle \lim\limits_{x \to 5}\frac{x^2-2x-15}{\sqrt{3+x}-5}\)
\(\displaystyle \lim\limits_{x \to 2}\frac{\sqrt{x^2+1}-\sqrt{5}}{x^2-4}\)
\(\displaystyle \lim\limits_{x \to 0}\frac{\sin(17x)}{42x}\)
\(\displaystyle \lim\limits_{x \to \infty}\frac{-x+\pi}{x^2+9x+42}\)
\(\displaystyle \lim\limits_{x \to -\infty}\frac{x^2+2x+1}{3x^2+3}\)
\(\displaystyle \lim\limits_{x \to -\infty}\frac{3x^2+x}{2x^2-9200}\)
\(\displaystyle \lim\limits_{x \to -\infty}\frac{3x^2+x^5}{2x^2-9200}\)
\(\displaystyle \lim\limits_{x \to \infty}\Bigr(3x^2-2x+1\Bigr)\)
\(\displaystyle \lim\limits_{x \to \infty}\frac{2x^2-32}{x^3-64}\)
\(\displaystyle \lim\limits_{x \to \infty} 7\)
\(\displaystyle \lim\limits_{x \to \infty}\frac{3x^2+4x}{x^4+2}\)
\(\displaystyle \lim\limits_{x \to -\infty}\frac{2x+3x^2+1}{2x^2+3}\)
\(\displaystyle \lim\limits_{x \to -\infty}\frac{x^3-3x^2+1}{3x^2+x+5}\)
\(\displaystyle \lim\limits_{x \to \infty}\frac{x^2+2}{x^3-2}\)
\(\displaystyle \lim\limits_{x \to \infty}\frac{\sin\bigl(\frac{1}{x}\bigr)}{x}\)
\(\displaystyle \lim\limits_{x \to -\infty}x^{2}\cos\biggl(\frac{1}{x}\biggr)\)
\(\displaystyle \lim\limits_{x \to \infty}\frac{\sin(\arctan(x))}{\arctan(-x)}\)