Estimating Limits from Graphs & Tables Quiz

1) Given a table of values for f(x) where x approaches 0:

x	f(x)
-0.1	0.1
-0.01	0.0001
-0.001	0.000001
0.001	0.000001
0.01	0.0001
0.1	0.01

What is the best estimate for lim(x→0) f(x)?

-0.1

0

0.01

0.1

Based on the table of values, the best estimate for lim(x→0) f(x) is 0. As x gets closer to 0 from both sides (values like -0.1, -0.01, -0.001 from the left and 0.001, 0.01, 0.1 from the right), f(x) gets closer to 0. The values approach 0 from above (0.01, 0.0001, 0.000001) on both sides, suggesting that 0 is the limiting value. Options A, C, and D are values that f(x) takes when x is farther from 0, not the limiting value.

Explanation

Based on the table of values, the best estimate for lim(x→0) f(x) is 0. As x gets closer to 0 from both sides (values like -0.1, -0.01, -0.001 from the left and 0.001, 0.01, 0.1 from the right), f(x) gets closer to 0. The values approach 0 from above (0.01, 0.0001, 0.000001) on both sides, suggesting that 0 is the limiting value. Options A, C, and D are values that f(x) takes when x is farther from 0, not the limiting value.

2) From a graph where the function approaches 2 from the left of x=0 but jumps to 4 on the right, what is limx→0 f(x)?

2

4

3

Does not exist

We assess the one-sided behaviors from the graph. The left-hand limit is 2, but the right-hand limit is 4. For the two-sided limit to exist, both must be equal. Since 2 ≠ 4, the limit does not exist.

Explanation

We assess the one-sided behaviors from the graph. The left-hand limit is 2, but the right-hand limit is 4. For the two-sided limit to exist, both must be equal. Since 2 ≠ 4, the limit does not exist.

3) Estimate limx→1 f(x) from these values: f(0.9)=1.9; f(0.99)=1.99; f(1.001)=2.001; f(1.01)=2.01.

1

2

1.99

Does not exist

We inspect the values around x=1. From left (0.9, 0.99), f approaches 2 (1.9 to 1.99). From right (1.001, 1.01), f approaches 2 (2.001 to 2.01). Both sides approach 2, so the limit is 2.

Explanation

We inspect the values around x=1. From left (0.9, 0.99), f approaches 2 (1.9 to 1.99). From right (1.001, 1.01), f approaches 2 (2.001 to 2.01). Both sides approach 2, so the limit is 2.

4) Estimate lim_{t→0} (e^t - 1)/t by using the following values: f(0.1) ≈1.10517; f(0.01) ≈1.01005; f(-0.1) ≈0.904837; f(-0.01) ≈0.99005.

0

1

E

Does not exist

We evaluate the table for (e^t - 1)/t near t=0. Positive side (0.1 to 0.01) approaches 1 (1.10517 to 1.01005). Negative side (-0.1 to -0.01) approaches 1 (0.904837 to 0.99005). Convergence from both sides to 1 gives the limit as 1.

Explanation

We evaluate the table for (e^t - 1)/t near t=0. Positive side (0.1 to 0.01) approaches 1 (1.10517 to 1.01005). Negative side (-0.1 to -0.01) approaches 1 (0.904837 to 0.99005). Convergence from both sides to 1 gives the limit as 1.

5) In a graph scaled to show x from -10 to 10, why might it miss the true limit behavior at x=0 for a function with rapid oscillations?

Oscillations are always visible

The broad scale hides fine details near x=0

Limits at infinity are affected instead

Scale only impacts unbounded functions

We consider how scale affects visibility. A wide x-range (-10 to 10) means each unit is small on the plot, so rapid changes or oscillations very close to x=0 may appear as a blur or solid line, missing the fact that the function doesn't settle to a value. Zooming in (narrower scale) reveals the true behavior for accurate limit estimation.

Explanation

We consider how scale affects visibility. A wide x-range (-10 to 10) means each unit is small on the plot, so rapid changes or oscillations very close to x=0 may appear as a blur or solid line, missing the fact that the function doesn't settle to a value. Zooming in (narrower scale) reveals the true behavior for accurate limit estimation.

6) For f(x) = cos(x) as x→0, estimate the limit from a graph showing the curve symmetric with respect to the y-axis, with a peak at (0,1).

0

1

-1

Does not exist

We observe the graph of cos(x), which at x=0 is 1, but near 0, it approaches 1 from both sides slightly below for small positive and negative x, getting closer. Both sides agree on 1, so the limit is 1.

Explanation

We observe the graph of cos(x), which at x=0 is 1, but near 0, it approaches 1 from both sides slightly below for small positive and negative x, getting closer. Both sides agree on 1, so the limit is 1.

7) Why does lim_{x→2} 1/(x-2)² not exist?

It approaches 0

It approaches +∞ from both sides, which is unbounded

It oscillates

Left and right differ

We analyze 1/(x-2)² near x=2. As x approaches 2 from left or right, (x-2)² is positive and small, so 1/(x-2)² becomes very large positive. Since it grows without bound (to +∞), the limit does not exist as a real number, due to unboundedness.

Explanation

We analyze 1/(x-2)² near x=2. As x approaches 2 from left or right, (x-2)² is positive and small, so 1/(x-2)² becomes very large positive. Since it grows without bound (to +∞), the limit does not exist as a real number, due to unboundedness.

8) Estimate limx→1 ln(x) by considering the following values: f(0.9)≈-0.0458, f(0.99)≈-0.00436, f(1.1)≈0.0414, f(1.01)≈0.0043

0

1

-∞

Does not exist

We use the values for ln(1+x) near x=0 (defined for x>-1). Positive side (0.1 to 0.01): 0.0953 to 0.00995, approaching 0. Negative side (-0.1 to -0.01): -0.1054 to -0.01005, approaching 0. Both sides to 0, limit is 0.

Explanation

We use the values for ln(1+x) near x=0 (defined for x>-1). Positive side (0.1 to 0.01): 0.0953 to 0.00995, approaching 0. Negative side (-0.1 to -0.01): -0.1054 to -0.01005, approaching 0. Both sides to 0, limit is 0.

9) Determine lim_{x→3} f(x) for f(x) = {x if x≤3, 6-x if x>3}.

Does not exist

3

6

0

We compute the one-sided limits. The left-hand limit lim_{x→3⁻} f(x) uses f(x)=x for x≤3, so approaches 3. The right-hand limit lim_{x→3⁺} f(x) uses f(x)=6-x for x>3, so approaches 6-3=3. Since both agree on 3, the limit is 3.

Explanation

We compute the one-sided limits. The left-hand limit lim_{x→3⁻} f(x) uses f(x)=x for x≤3, so approaches 3. The right-hand limit lim_{x→3⁺} f(x) uses f(x)=6-x for x>3, so approaches 6-3=3. Since both agree on 3, the limit is 3.

10) For exponential f(x)=2^x, estimate lim_{x→0} 2^x using graph approaching 1.

0

1

2

∞

We observe the graph of 2^x, which at x=0 is 1, and near 0 approaches 1 from below on left, above on right, but both sides to 1. The limit is 1.

Explanation

We observe the graph of 2^x, which at x=0 is 1, and near 0 approaches 1 from below on left, above on right, but both sides to 1. The limit is 1.

11) Why does lim_{x→0} x sin(1/x) exist even with oscillation in sin(1/x)?

DNE due to oscillation

Exists as 0, since |x sin(1/x)| ≤ |x| →0

1

Unbounded

We note that sin(1/x) is bounded by -1 ≤ sin(1/x) ≤1, so -|x| ≤ x sin(1/x) ≤ |x|. As x→0, both -|x| and |x| approach 0. By the squeeze theorem, since the function is squeezed between two functions approaching 0, the limit is 0.

Explanation

We note that sin(1/x) is bounded by -1 ≤ sin(1/x) ≤1, so -|x| ≤ x sin(1/x) ≤ |x|. As x→0, both -|x| and |x| approach 0. By the squeeze theorem, since the function is squeezed between two functions approaching 0, the limit is 0.

12) Find lim_{x→1} log10(x) using the following table of values: x | log10(x) 0.9 | ≈-0.0458 0.99 | ≈-0.00436 1.01 | ≈0.0043 1.1 | ≈0.0414

0

1

-∞

DNE

We analyze the table for log10(x) near x=1. From left (0.9 to 0.99), values approach 0 from negative (-0.0458 to -0.00436). From right (1.01 to 1.1), approach 0 from positive (0.0043 to 0.0414, but closer values would be smaller). Both sides approach 0, so the limit is 0.

Explanation

We analyze the table for log10(x) near x=1. From left (0.9 to 0.99), values approach 0 from negative (-0.0458 to -0.00436). From right (1.01 to 1.1), approach 0 from positive (0.0043 to 0.0414, but closer values would be smaller). Both sides approach 0, so the limit is 0.

13) Find lim(t→0) sin(t)/t by considering the following values: t=0.2, f(t)≈0.9933 t=0.02, f(t)≈0.999933 t=-0.2, f(t)≈0.9933 t=-0.02, f(t)≈0.999933

0

1

Pi

DNE

We review the table for sin(t)/t near t=0. Positive side (0.2 to 0.02) approaches 1 (0.9933 to 0.999933). Negative side (-0.2 to -0.02) approaches 1 (same by even-odd properties). Both sides to 1, limit is 1.

Explanation

We review the table for sin(t)/t near t=0. Positive side (0.2 to 0.02) approaches 1 (0.9933 to 0.999933). Negative side (-0.2 to -0.02) approaches 1 (same by even-odd properties). Both sides to 1, limit is 1.

14) Which notation correctly represents the limit of (x³ - 8)/(x-2) as x approaches 2?

Limx→2 (x³ - 8)/(x-2)

Limx=2 (x³ - 8)/(x-2)

Limx→2 (2³ - 8)/(2-2)

Limx→2+ (x³ - 8)/(x-2) only

We start by identifying the need to express the limit analytically with the correct notation. The function is (x³ - 8)/(x-2), so the limit as x approaches 2 is written as lim_{x→2} of the expression. The arrow → indicates "approaches" rather than equals, and we do not evaluate at x=2 since the limit concerns values near 2, not at 2. Thus, the correct notation is lim_{x→2} (x³ - 8)/(x-2).

Explanation

We start by identifying the need to express the limit analytically with the correct notation. The function is (x³ - 8)/(x-2), so the limit as x approaches 2 is written as lim_{x→2} of the expression. The arrow → indicates "approaches" rather than equals, and we do not evaluate at x=2 since the limit concerns values near 2, not at 2. Thus, the correct notation is lim_{x→2} (x³ - 8)/(x-2).

15) What does lim_{x→c} f(x) = R imply about f(x) near x=c?

F(c)=R always

F(x) can be made arbitrarily close to R by x close to c, not equal

F(x) is constant R

Only one-sided approach matters

B. f(x) can be made arbitrarily close to R by x close to c, not equal

We begin by recalling the definition of a limit. The limit lim_{x→c} f(x) = R means that for any small positive number epsilon, there exists a delta such that if 0 < |x - c| < delta, then |f(x) - R| < epsilon. This means as x approaches c from values not equal to c, f(x) can be made as close to R as desired by choosing x sufficiently close to c. This interpretation does not require f(c) to be defined or equal to R, and it applies from both sides unless specified otherwise.

Explanation

We begin by recalling the definition of a limit. The limit lim_{x→c} f(x) = R means that for any small positive number epsilon, there exists a delta such that if 0 < |x - c| < delta, then |f(x) - R| < epsilon. This means as x approaches c from values not equal to c, f(x) can be made as close to R as desired by choosing x sufficiently close to c. This interpretation does not require f(c) to be defined or equal to R, and it applies from both sides unless specified otherwise.