Panhellenic Exams · 90 minutes · 40 marks
Chapter 1: Limits and continuity · Chapter 3: Integral calculus
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Find lim(x→6) (x²−36)/(x−6).
The form 0/0 is not an answer: factor first.
(x²−36)/(x−6)=x+6, x≠6
As x→6, x+6→12.
12
For x<0, f(x)=x+7; for x≥0, f(x)=6x+8. Does a limit exist at 0?
The left-hand limit is 7.
The right-hand limit is 8.
The one-sided limits differ.
There is neither a two-sided limit nor continuity at 0.
Show that g(x)=x³+x−7 has a root in (0,7).
g is a polynomial, so it is continuous on the closed interval.
g(0)=−7<0; g(7)=343>0
The intermediate value theorem guarantees at least one root in the open interval.
At least one root exists; that is all this argument establishes.
If f(7)=6, can that fact alone establish lim(x→7)f(x)=6?
A limit concerns nearby values, not only the value at the point.
If f(x)=0 for x≠7 and f(7)=6, the limit is 0.
This is a counterexample to the claim.
Not without additional information, such as continuity.
Evaluate the definite integral of f(x)=2x+7 from x=0 to x=6.
Choose antiderivative F(x)=x²+7x.
F(6)−F(0)=36+42−0
Differentiating F gives the integrand.
78
Find the area between y=7x and y=6x for 0≤x≤2.
For x≥0 the first line is above the second. Integrate the difference from 0 to 2.
F(x)=(7−6)x²/2=x²/2; E=F(2)−F(0)=2²/2−0=2
The difference is non-negative on the stated interval.
2
For f(x)=x on [−6,6], calculate the definite integral and total geometric area.
F(x)=x²/2; F(6)−F(−6)=6²/2−(−6)²/2=0
Each triangle has area 6·6/2=36/2.
Geometric areas add: 36/2+36/2=36.
Integral 0; area 36.
f(x)=−7 on [0,6]. Is its area equal to the integral? Calculate both.
F(x)=−7x; F(6)−F(0)=−7·6−0=−42
The graph lies below the x-axis.
E=|−42|=42
No: integral −42, area 42.
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