Graphing Functions and Finding Limits for $ |x + 3| $, $ e^{-x} $, $ 2 + \ln x $

Question

Graph $ f(x) = |x + 3| $, $ g(x) = e^{-x} $, $ h(x) = 2 + \ln x $ and find the following limits: $ \lim_{x \to -3} f(x) $, $ \lim_{x \to 0} f(x) $, $ \lim_{x \to -\infty} g(x) $, $ \lim_{x \to \infty} g(x) $, $ \lim_{x \to 0^+} h(x) $, $ \lim_{x \to \frac{1}{e}} h(x) $

Accepted Answer

Limits: $ 0 $, $ 3 $, $ \infty $, $ 0 $, $ -\infty $, $ 1 $ (respectively)

Answer

1. For $ \lim_{x \to -3} f(x) $: $ f(x) = |x + 3| $ is continuous at $ x = -3 $, so the limit equals $ f(-3) = |-3 + 3| = 0 $. 2. For $ \lim_{x \to 0} f(x) $: $ f(x) = |x + 3| $ is continuous at $ x = 0 $, so the limit equals $ f(0) = |0 + 3| = 3 $. 3. For $ \lim_{x \to -\infty} g(x) $: $ g(x) = e^{-x} = e^{|x|} $. As $ x \to -\infty $, $ |x| \to \infty $, so $ e^{|x|} \to \infty $. 4. For $ \lim_{x \to \infty} g(x) $: $ g(x) = e^{-x} = \frac{1}{e^x} $. As $ x \to \infty $, $ e^x \to \infty $, so $ \frac{1}{e^x} \to 0 $. 5. For $ \lim_{x \to 0^+} h(x) $: $ h(x) = 2 + \ln x $. As $ x \to 0^+ $, $ \ln x \to -\infty $, so $ 2 + \ln x \to -\infty $. 6. For $ \lim_{x \to \frac{1}{e}} h(x) $: $ h(x) $ is continuous at $ x = \frac{1}{e} $ (logarithmic functions are continuous on their domain $ x > 0 $). Thus, the limit equals $ h\left( \frac{1}{e} \right) = 2 + \ln\left( \frac{1}{e} \right) = 2 - 1 = 1 $.