Free Math Worksheets

- Check that each expression is defined at the approach value. - Then substitute directly and simplify carefully.

1. Substitute \(x=2\): \(\frac{3\cdot 2^2-4}{2+2}=\frac{8}{4}=2\). 2. Substitute \(x=9\): \(\sqrt{9}\cdot(9-7)=3\cdot 2=6\). 3. Substitute \(x=-1\): \((-1)^5+3\cdot(-1)^2-1=-1+3-1=1\).

1) \(2\) 2) \(6\) 3) \(1\)

- Rewrite quotients when possible. - Recall the behavior of exponential functions with bases greater than \(1\) and between \(0\) and \(1\).

1. In part a, \(\frac{4}{x^2}\to0\) in both directions, so \(f(x)\to5\). 2. Rewrite part b as \(g(x)=3+\frac{2}{4^x}\). As \(x\to\infty\), the fraction approaches \(0\), so \(g(x)\to3\). As \(x\to-\infty\), \(4^x\to0^+\), so the fraction approaches \(+\infty\), and \(g(x)\to+\infty\). 3. In part c, \((0.5)^x\to0\) as \(x\to\infty\), so \(h(x)\to-2\). As \(x\to-\infty\), \((0.5)^x\to+\infty\), so \(h(x)\to+\infty\).

a) \(\lim_{x\to\pm\infty}f(x)=5\) b) \(\lim_{x\to\infty}g(x)=3\); \(\lim_{x\to-\infty}g(x)=+\infty\) c) \(\lim_{x\to\infty}h(x)=-2\); \(\lim_{x\to-\infty}h(x)=+\infty\)