Recall some limit properties…
• limits distribute over sums:
[tex]\displaystyle \lim_{x\to c}(f(x)+g(x)) = \lim_{x\to c}f(x)+\lim_{x\to c}g(x)[/tex]
• limits distribute over products:
[tex]\displaystyle \lim_{x\to c}(f(x)\times g(x)) = \lim_{x\to c}f(x)\times\lim_{x\to c}g(x)[/tex]
• limits distribute over quotients, provided that the denominator doesn't approach 0 :
[tex]\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)} = \frac{\displaystyle\lim_{x\to c}f(x)}{\displaystyle\lim_{x\to c}g(x)}[/tex]
• if f(x) is continuous at x = c, then the limit "passes through" a composition:
[tex]\displaystyle \lim_{x\to c}f(g(x)) = f\left(\lim_{x\to c}g(x)\right)[/tex]
# # #
(a) This limit is 2. At x = 2, we have f (2) = 1, but from either side of x = 2, we see f(x) approaching the point (2, 2). So
[tex]\displaystyle \lim_{x\to 2}(f(x)+g(x)) = \lim_{x\to2}f(x)+\lim_{x\to2}g(x) = 2+0 = 2[/tex]
(b) This limit does not exist. We would have
[tex]\displaystyle \lim_{x\to1}(2f(x)-3g(x)) = 2\lim_{x\to1}f(x)-3\lim_{x\to1}g(x)[/tex]
but g(x) approaches 2 from the left of x = 1, and g(x) approaches 1 from the right of x = 1. The one-sided limits don't match, so the two-sided limit doesn't exist.
(c) This limit is 0. It looks like f(x) passes through the origin, while g(x) ≈ 3/2 at x = 0. So
[tex]\displaystyle\lim_{x\to0}f(x)g(x) = \lim_{x\to0}f(x)\times\lim_{x\to0}g(x)=0\times\frac32 = 0[/tex]
(d) This limit does not exist since
[tex]\displaystyle \lim_{x\to-1}g(x)=0[/tex]
(e) This limit is 16. Nothing tricky here, just use the same property as in (c).
[tex]\displaystyle\lim_{x\to2}x^3f(x) = \lim_{x\to2}x^3\times\lim_{x\to2}f(x) = 8\times2=16[/tex]
(f) This limit is 1. f(x) is continuous at x = 1, while g(x) approaches 2 from the left.
[tex]\displaystyle\lim_{x\to1^-}f(g(x)) = f\left(\lim_{x\to1^-}g(x)\right) = f(2) = 1[/tex]
