Answer:
Use continuity to evaluate the limit.
lim 16+radical x/ radical 16+x
1+9
Consider the intervals for which the numerator and the denominator are continuous.
The numerator 16+ radical x is contintuous on the interval
[0,00)
0
The denominator V16 + I is continuous and nonzero on the interval
(16,00)
X
The interval for which the quotient is continuous is the intersection of the above intervals.
16+
Therefore, the quotient
is continuous on the interval
716 +1
[-16,0]
X
Since 2
9 is in the interval (0, 0), then fis continuous at x = 9. Therefore,
19
16 + VE
lim
179 16 +2
f(9)
00
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