Answer
a) $x$ belngs to the open interval $(-0.1, 0.1)$.
b) $x$ belongs to the open interval: $(2.9975, 3.0025)$.
c) $x$ belongs to the open interval: $(\sqrt {15.999}, \sqrt {16.001}) $
Work Step by Step
a) We have: $\epsilon = 0.1$
$f(x) = x + 2$
$f(0) = 2$
$x$ is centered around 0: thus, $f(0) = L = 2$
Using the definition of limits, we have: $|f(x) - L| < \epsilon$
Thus, $|x + 2 - 2| < 0.1$
Thus, $|x| < 0.1$.
Thus, all x's in the open interval $(-0.1, 0.1)$ satisfy this condition.
Thus, $-0.1 < x <0.1$
b) We have: $\epsilon = 0.01$
$f(x) = 4x - 5$
$f(3) = 7$
$x$ is centered around 3: thus, $f(3) = L = 7$
Using the definition of limits, we have: $|f(x) - L| < \epsilon$
Thus, $|4x - 5 - 7| < 0.01$
Thus, $|4x - 12| < 0.01$
Thus, $|4(x - 3)| < 0.01$
Thus, $|x - 3| < \frac{0.01}{4}$
Expanding the inequality above, we get: $3 - \frac{0.01}{4} < x < 3 + \frac{0.01}{4}$
Thus, $x$ belongs to the open interval: $(2.9975, 3.0025)$.
c) We have: $\epsilon = 0.001$
$f(x) = x^{2}$
$f(4) = 16$
$x$ is centered around 4: thus, $f(4) = L = 16$
Using the definition of limits, we have: $|f(x) - L| < \epsilon$
Thus, $|x^{2} - 16| < 0.001$
Thus, $16 - 0.001 < x^{2} < 16 + 0.001$
Thus, $15.999 < x^2 < 16.001$
Taking the root on both sides of the inequality,
$\sqrt {15.999} < x < \sqrt {16.001}$
Thus, $x$ belongs to the open interval: $(\sqrt {15.999}, \sqrt {16.001}) $
