Question 1065623
Complete the table by computing f(x) at the given values of x. (Round your answers to three decimal places.)

f(x) = 2x^2 - 4

x	3.9	3.99	3.999	4.001	4.01	4.1
f(x)	 
<pre><b>
f(3.9) = 2(3.9)^2-4 = 2(15.21)-4 = 30.42-4 = 26.42
f(3.99) = 2(3.99)^2-4 = 2(15.9201)-4 = 31.8402-4 = 27.8402
f(3.999) = 2(3.999)^2-4 = 2(15.992001)-4 = 31.984002-4 = 27.984002
f(4.001) = 2(4.001)^2-4 = 2(16.008001)-4 = 32.016002-4 = 28.016002
f(4.01) = 2(4.01)^2-4 = 2(16.0801)-4 = 32.1602-4 = 28.1602
f(4.1) = 2(4.1)^2-4 = 2(16.81)-4 = 33.62-4 = 29.62

x	3.9	3.99	3.999	4.001	4.01	4.1
f(x)   26.42   27.840  27.984  28.016  28.160  29.62  
</pre>
Use these results to estimate the indicated limit (if it exists). 
<pre>
When x is a tiny bit less than 4, say 3.999, f(x) is 27.984
When x is a tiny bit more than 4, say 4.001, f(x) is 28.016

So when x is very close to 4, f(x) is very close to 28.

lim f(x)     = 28
x &#8594; 4 

[You may wonder why not just plug x=4 in f(x) = 2x^2-4 and
get f(4) = 2(4)^2-4 = 2(16)-4 = 32-4 = 28?  It's because
sometimes you can't just plug in the number and get the limit.
This one you can, but your teacher wants you to get used to
the idea of only plugging in numbers close to what x is approaching
without actually plugging in what x is approaching. Then when 
you get to functions that you can't just plug in the number x
is approaching, you'll know what to do.]

Edwin</pre><b>