Answer to Question #226666 in Calculus for Portia

Question

A). Determine the following limits (if they exist)

1. Lim X approaches -5 x^2+x-20/3(X+5)

2. Lim t approaches 0 sin5t/t^2+4t

3. Lim approaches -infinity 3-|X|/2|X|+1

4. Lim X approaches 0 2x/3-√x+9

5. Lim X approaches +infinity 2x+x^2+1/1-x+2x^2

Answer 1

$\displaystyle 1.\\ \begin{aligned} \lim_{x \to -5} \frac{x^2+x-20}{3(x+5)} &= \lim_{x \to -5} \frac{(x + 5)(x - 4)}{3(x+5)} \\&= \lim_{x \to -5} \frac{x - 4}{3} \\&= \frac{-9}{3} = -3 \end{aligned}\\ 2.\\ \begin{aligned} \lim_{x \to 0} \frac{\sin(5t)}{t^2+4t} &= \lim_{x \to 0} \frac{5\cos(5t)}{2t + 4} \\&= \frac{5}{4} \end{aligned}\\ 3.\\ \begin{aligned} \lim_{x \to -\infty} \frac{3-|x|}{2|x|+1} &= \lim_{x \to -\infty} \frac{\frac{3}{|x|} - 1}{2+\frac{1}{|x|}} \\&= \frac{-1}{2} \end{aligned}\\ 4.\\ \begin{aligned} \lim_{x \to 0} \frac{2x}{3 - \sqrt{x} + 9} &= \frac{2(0)}{3 - \sqrt{0} + 9} \\&= \frac{0}{12} = 0 \end{aligned}\\ 5.\\ \begin{aligned} \lim_{x \to \infty} \frac{2x+x^2+1}{1-x+2x^2} &= \lim_{x \to \infty} \frac{2/x + 1 + 1/x^2}{1/x^2 - 1/x +2} \\&= \frac{1 + 0 + 0}{2 - 0 + 0} = \frac{1}{2} \end{aligned}$