Answer to Question #246052 in Calculus for Finn

Question

Question No#13 Let h(𝑥) = 𝑥2−2𝑥−3

a) Make a table of the values of h(x) at x = 2.9, 2.99, 2.999, and so on. Then estimate limh(𝑥). What estimate do you arrive at if you evaluate h at x = 3.1, 3.01, 3.001 ...

𝑥→3

instead?

b) Support your conclusions in part (a) by graphing h near x = 3 and using that graph to

estimate h(x) on the graph as x approaching 3.

c) Find lim h(𝑥) algebraically.

𝑥→3

Answer 1

a)

$f(2.9)=-0.39,\ f(2.99)=-0.0399,\ f(2.999)=-0.003999,$

$f(2.9999)=-0.00039999$

$\displaystyle{\lim_{x\to 3^-}}(x^2-2x-3)=0$

$f(3.1)=0.41,\ f(3.01)=0.0401,\ f(3.001)=0.004001,$

$f(3.0001)=0.00040001$

$\displaystyle{\lim_{x\to 3^+}}(x^2-2x-3)=0$

b)

c)

$\displaystyle{\lim_{x\to 3^-}}(x^2-2x-3)=3^2-2\cdot 3-3=0$