Evaluating Limits using Limit Theorems
Limits
Evaluate the following limits. Use limit theorems, not ε – δ techniques. If any of them fail to exist, say so and say why.
| 1. (a) limx→10 x 2 − 100 x − 10
| (b) limx→10 x 2 − 99 x − 10 | (c) limx→10 x 2 − 100 x − 9 | (d) limx→10 f(x), where f(x) = x 2 for all x 6= 10, but f(10) = 99. | (e) limx→10 √ −x 2 + 20x − 100 |
| 2. limx→−4 x 2 − 16 x + 4 ln |x| | 3. limx→∞ x 2 e 4x − 1 − 4x
| 4. lim x→−∞ 3x 6 − 7x 5 + x 5x 6 + 4x 5 − 3 | 5. lim x→−∞ 5x 7 − 7x 5 + 1 2x 7 + 6x 6 − 3 | 6. lim x→−∞ 2x + 3x 3 x 3 + 2x – 1 |
| 7. lim x→−∞ 5x + 2x 3 x 3 + x − 7
| 8. lim x→−∞ 3x + |1 − 3x| 1 − 5x
| 9. limu→∞ u √ u 2 + 1 | 10. limx→∞ 1 + 3x √ 2x 2 + x | 11. limx→∞ √ 4x 2 + 3x − 7 7 − 3x |
| 12. lim x→1+ √ x − 1 x 2 − 1
| 13. Let f(x) = ( x 2−1 |x−1| if x 6= 1, 4 if x = 1. Find Lim x→1− f(x). | 14. . Let F(x) = 2x 2−3x |2x−3| . (a) Find lim x→1.5+ F(x). (b) Find lim x→1.5− F(x). (c) Does limx→1.5 F(x) exist? Provide a reason. | 15. limx→8 (x − 8)(x + 2 |
To evaluate the given limits, we will directly substitute the values into the expressions. If the result is of the form “∞” or “−∞,” then the limit is infinite. If we obtain an indeterminate form, such as 0/0 or ∞/∞, we will apply L’Hôpital’s rule to further evaluate the limit.
(a) lim(x→1−) (x + 1)/(x^2 − 1):
As x approaches 1 from the left (x → 1-), the expression becomes 0/0, which is an indeterminate form. We can use L’Hôpital’s rule:
lim(x→1-) (x + 1)/(x^2 − 1) = lim(x→1-) d/dx(x + 1)/d/dx(x^2 − 1) = lim(x→1-) 1/(2x) = 1/(2*1) = 1/2
(b) lim(x→0) sin(6x)/(2x):
As x approaches 0, the expression becomes 0/0, which is an indeterminate form. Again, we can use L’Hôpital’s rule:
lim(x→0) sin(6x)/(2x) = lim(x→0) d/dx(sin(6x))/d/dx(2x) = lim(x→0) 6cos(6x)/2 = 6cos(0)/2 = 6/2 = 3
(c) lim(x→0) sinh(2x)/(x*ex):
As x approaches 0, the expression becomes 0/0, which is an indeterminate form. We can use L’Hôpital’s rule:
lim(x→0) sinh(2x)/(xex) = lim(x→0) d/dx(sinh(2x))/d/dx(xex) = lim(x→0) 2cosh(2x)/(ex + xex) = 2*cosh(0)/(1 + 0) = 2/1 = 2
(d) lim(x→0+) (x/(0.01*ln(x))):
As x approaches 0 from the right (x → 0+), the expression becomes 0/(0.01*ln(0)), which is of the form 0/−∞. This is still an indeterminate form, so we can apply L’Hôpital’s rule:
lim(x→0+) (x/(0.01ln(x))) = lim(x→0+) d/dx(x)/d/dx(0.01ln(x)) = lim(x→0+) 1/(0.01*(1/x)) = lim(x→0+) 1/(0.01/x) = lim(x→0+) x/0.01 = 0/0.01 = 0
Therefore, the evaluated limits are: (a) lim(x→1-) (x + 1)/(x^2 − 1) = 1/2 (b) lim(x→0) sin(6x)/(2x) = 3 (c) lim(x→0) sinh(2x)/(xex) = 2 (d) lim(x→0+) (x/(0.01ln(x))) = 0
