Question #246052

Question No#13 Let h(š‘„) = š‘„2āˆ’2š‘„āˆ’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(š‘„). What estimate do you arrive at if you evaluate h at x = 3.1, 3.01, 3.001 ...

š‘„ā†’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(š‘„) algebraically.

š‘„ā†’3



Expert's answer

a)

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

f(2.9999)=āˆ’0.00039999f(2.9999)=-0.00039999

lim⁔x→3āˆ’(x2āˆ’2xāˆ’3)=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

lim⁔x→3+(x2āˆ’2xāˆ’3)=0\displaystyle{\lim_{x\to 3^+}}(x^2-2x-3)=0


b)



c)

lim⁔x→3āˆ’(x2āˆ’2xāˆ’3)=32āˆ’2ā‹…3āˆ’3=0\displaystyle{\lim_{x\to 3^-}}(x^2-2x-3)=3^2-2\cdot 3-3=0




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS