Answer
(a) Since the values are getting very small as $x$ approaches 0, we could guess that $\lim\limits_{x \to 0} (x^2-\frac{2^x}{1000}) = 0$
(b) The values are getting close to $-0.001$ as $x$ approaches 0 more closely. Therefore, we could guess that $\lim\limits_{x \to 0} (x^2-\frac{2^x}{1000}) = -0.001$
Work Step by Step
(a) We can evaluate $f(x) = x^2-(\frac{2^x}{1000})$ for values of $x$ that approach $0$:
$f(1) = (1)^2-(\frac{2^{1}}{1000}) = 0.998$
$f(0.8) = (0.8)^2-(\frac{2^{0.8}}{1000}) = 0.63826$
$f(0.6) = (0.6)^2-(\frac{2^{0.6}}{1000}) = 0.35848$
$f(0.4) = (0.4)^2-(\frac{2^{0.4}}{1000}) = 0.15868$
$f(0.2) = (0.2)^2-(\frac{2^{0.2}}{1000}) = 0.03885$
$f(0.1) = (0.1)^2-(\frac{2^{0.1}}{1000}) = 0.008928$
$f(0.05) = (0.05)^2-(\frac{2^{0.05}}{1000}) = 0.0014647$
Since the values are getting very small as $x$ approaches 0, we could guess that $\lim\limits_{x \to 0} (x^2-\frac{2^x}{1000}) = 0$
(b) We can evaluate $f(x) = x^2-(\frac{2^x}{1000})$ for more values of $x$ that approach $0$ more closely:
$f(0.04) = (0.04)^2-(\frac{2^{0.04}}{1000}) = 0.000572$
$f(0.02) = (0.02)^2-(\frac{2^{0.02}}{1000}) = -0.000614$
$f(0.01) = (0.01)^2-(\frac{2^{0.01}}{1000}) = -0.000907$
$f(0.005) = (0.005)^2-(\frac{2^{0.005}}{1000}) = -0.0009785$
$f(0.003) = (0.003)^2-(\frac{2^{0.003}}{1000}) = -0.000993$
$f(0.001) = (0.001)^2-(\frac{2^{0.001}}{1000}) = -0.0009997$
The values are getting close to $-0.001$ as $x$ approaches 0 more closely. Therefore, we could guess that $\lim\limits_{x \to 0} (x^2-\frac{2^x}{1000}) = -0.001$
