Answer
a) it appears in $ \lim\limits_{x \to 0}=0$
b) it appears in $ \lim\limits_{x \to 0}=-0.001$
Work Step by Step
Given:
$f(x)=x^2-\frac{2^x}{1000}$
a)
Let suppose $x=1,0.8,0.6,0.4,0.2,0.1,0.05$
for $x=1$
$f(1)=1^2-\frac{2^1}{1000}$
$=1-\frac{2}{1000}$
$=\frac{1000-2}{1000}$
$=\frac{998}{1000}$
$=0.998$
for $x=0.8$
$f(0.8)=(0.8)^2-\frac{2^{0.8}}{1000}$
$f(0.8)=0.638$
for $x=0.6$
$f(0.6)=(0.6)^2-\frac{2^{0.6}}{1000}$
$f(0.6)=0.358$
for $x=0.4$
$f(0.4)=(0.4)^2-\frac{2^{0.4}}{1000}$
$f(0.4)=0.158$
for $x=0.2$
$f(0.2)=(0.2)^2-\frac{2^{0.2}}{1000}$
$f(0.2)=0.038$
for $x=0.1$
$f(0.1)=(0.1)^2-\frac{2^{0.1}}{1000}$
$f(0.1)=0.008$
for $x=0.05$
$f(0.05)=(0.05)^2-\frac{2^{0.05}}{1000}$
$f(0.05)=0.001$
hence, it appears that $\lim\limits_{x \to 0}=0$
Similarly,
b)
Let suppose $x=0.04,0.02,0.01,0.005,0.003,0.001$
for $x=0.04$
$f(0.04)=(0.04)^2-\frac{2^{0.04}}{1000}$
$f(0.8)=0.000 $
for $x=0.02$
$f(0.02)=(0.02)^2-\frac{2^{0.02}}{1000}$
$f(0.02)=−0.000614$
for $x=0.01$
$f(0.01)=(0.01)^2-\frac{2^{0.01}}{1000}$
$f(0.01)=−0.000 907$
for $x=0.005$
$f(0.005)=(0.005)^2-\frac{2^{0.005}}{1000}$
$f(0.005)=−0.000 978$
for $x=0.003$
$f(0.003)=(0.003)^2-\frac{2^{0.003}}{1000}$
$f(0.003)=−0.000 993$
for $x=0.001$
$f(0.001)=(0.001)^2-\frac{2^{0.001}}{1000}$
$f(0.1)=−0.001 000$
Hence, it appears in $ \lim\limits_{x \to 0}=-0.001$
