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Sagot :
Answer:
[tex]\lim_{t \to \ 16} (t-16)(t-4) =0[/tex]
Step-by-step explanation:
g(t)= (t-16)(t-4)
(a) Then, we have,
t g(t)
15.9 -1.19
15.99 -0.1199
15.999 -0.011999
Also, we have,
t g(t)
16.1 1.21
16.01 0.1201
16.001 0.012001
(b) We can see, from the above table that as 't' tends to 16 from the left hand side, the value of g(t)= (t-16)(t-4) is tending to 0 from the negative direction.
Similarly, from the second table, we can see that as 't' tends to 16 from the right hand side, the value of g(t)= (t-16)(t-4) is tending to 0 from the positive direction.
Thus, we can say that [tex]\lim_{t \to \ 16-} (t-16)(t-4)[/tex] & [tex]\lim_{t \to \ 16+} (t-16)(t-4)[/tex], i.e., the left hand limit and right hand limit of g(t)= (t-16)(t-4), at t=16, both exist and is both equal to 0.
Since both these limits exist and are equal, we can say, [tex]\lim_{t \to \ 16-} (t-16)(t-4) = \lim_{t \to \ 16+} (t-16)(t-4) = \lim_{t \to \ 16} (t-16)(t-4) = 0[/tex]
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