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(a) Evaluate the function $ f(x) = x^2 - (2^x/1000) $ for $ x $ = 1, 0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of $$ \lim_{x \to 0}\left( x^2 - \frac{2^x}{1000} \right) $$

(b) Evaluate $ f(x) $ for $ x $ = 0.04, 0.02, 0.01, 0.005, 0.003, and 0.001. Guess again.

$$\text { For } f(x)=x^{2}-\left(2^{x} / 1000\right)$$

$$\begin{aligned}

&(a)\\

&\begin{array}{|l|c|}

\hline x & f(x) \\

\hline 1 & 0.998000 \\

0.8 & 0.638259 \\

0.6 & 0.358484 \\

0.4 & 0.158680 \\

0.2 & 0.038851 \\

0.1 & 0.008928 \\

0.05 & 0.001465 \\

\hline

\end{array}

\end{aligned}$$

$$\text { It appears that } \lim _{x \rightarrow 0} f(x)=0$$

(b) $$\begin{aligned}

&\begin{array}{|c|r|}

\hline x & f(x) \\

\hline 0.04 & 0.000572 \\

0.02 & -0.000614 \\

0.01 & -0.000907 \\

0.005 & -0.000978 \\

0.003 & -0.000993 \\

0.001 & -0.001000 \\

\hline

\end{array}\\

&\text { It appears that } \lim _{x \rightarrow 0} f(x)=-0.001

\end{aligned}$$

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This is problem number forty nine of Stuart Calculus Aches Edition Section two point two, Party says evaluate the function of X equals X squared, minus the quantity to the ex tried by a thousand four x equals one zero point e their point six zero point four there a point to you, Sarah point one and zero point zero five and then gets the value of this limited, which is eliminated. Function as X approaches. Zero. Just a quick note regarding the values that are given. It seems as if the problem is only asking for the limit. Has expressed zero from the right since we're only using these X values and now to set their only positive values. So really, this solution is for the limit as experts zero from the right. Now we take these valleys and we evaluate dysfunction. And what we're trying to achieve is we're trying to estimate the limit as explosions here from the right by a protein zero from one two zero point eight deserved Win six and getting closer to zero, and we've only with their function and see what the pattern. Maybe once we reach point five this at the crime state at the current point for party is our best estimate. The best estimate that we can give for this limit is going to be a point zero zero one for six. So we're going to write that out point zero zero one four six and that is there solution party From what we were given already from what we were given for Part B, we're going Tio Junction further getting closer to zero by using the X values point for point out too. Point one point oh, five point o three and finally point o O one. So we extend our calculations of it and we see that the value continues to change until about these last two points where we see that the limit or the function converges at a single point, it is approximately negative zero point zero zero one. So we're going tow more definitively states at the solution to the limit is negative zero point zero zero one and that is their second guess. And that is something we feel more confident about. Quick. What? The limit of this function as expert is here from the right that this is a mark, correct. Guess then the first guests