Question #246052

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



1
Expert's answer
2021-10-06T10:21:07-0400

a)

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

f(2.9999)=0.00039999f(2.9999)=-0.00039999

limx3(x22x3)=0\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(3.1)=0.41,\ f(3.01)=0.0401,\ f(3.001)=0.004001,

f(3.0001)=0.00040001f(3.0001)=0.00040001

limx3+(x22x3)=0\displaystyle{\lim_{x\to 3^+}}(x^2-2x-3)=0


b)



c)

limx3(x22x3)=32233=0\displaystyle{\lim_{x\to 3^-}}(x^2-2x-3)=3^2-2\cdot 3-3=0




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