Evaluate the limit, if it exists. lim (h-9)²-81 h→0 h Step 1 We are asked to find the following limit. lim (h-9)² - 81 h 0 h We will first attempt to do so using direct substitution. Let f(h) = (h- 9)²-81. We know that as f is a rational function, if 0 is in the domain off, then the given limit is equal to f(0). The function f is not h is not defined at h = 0; therefore, we cannot evaluate the limit by this direct substitution. Therefore, we will next attempt to use the quotient law, which we recall is as follows. lim g(x) h-0 lim g(x) h→0 k(x) We note that the quotient law requires that the limit in the denominator is not equal to 0, so we must check that as follows. lim h = 0✔ 0 h→0 So, this means that the quotient law cannot be used ✔ (h - 9)²-81 - 9)² - h = = lim k(x) h→0 Step 2 We have observed that we cannot use direct substitution or the quotient law to find the limit. We can now attempt to simplify the given function before evaluating the limit to see if that works. We note that if we expand the numerator and then factor it, the resulting rational expression can be simplified as follows. h²_ x )h +81-81 )n_h(h-([ h² =h- h cannot be used h h Note that it is valid to cancel the common factor h from the numerator and denominator because h = 0 for the values the function is evaluated at for the limit. The variable h approaches 0 but is not equal to 0.
Evaluate the limit, if it exists. lim (h-9)²-81 h→0 h Step 1 We are asked to find the following limit. lim (h-9)² - 81 h 0 h We will first attempt to do so using direct substitution. Let f(h) = (h- 9)²-81. We know that as f is a rational function, if 0 is in the domain off, then the given limit is equal to f(0). The function f is not h is not defined at h = 0; therefore, we cannot evaluate the limit by this direct substitution. Therefore, we will next attempt to use the quotient law, which we recall is as follows. lim g(x) h-0 lim g(x) h→0 k(x) We note that the quotient law requires that the limit in the denominator is not equal to 0, so we must check that as follows. lim h = 0✔ 0 h→0 So, this means that the quotient law cannot be used ✔ (h - 9)²-81 - 9)² - h = = lim k(x) h→0 Step 2 We have observed that we cannot use direct substitution or the quotient law to find the limit. We can now attempt to simplify the given function before evaluating the limit to see if that works. We note that if we expand the numerator and then factor it, the resulting rational expression can be simplified as follows. h²_ x )h +81-81 )n_h(h-([ h² =h- h cannot be used h h Note that it is valid to cancel the common factor h from the numerator and denominator because h = 0 for the values the function is evaluated at for the limit. The variable h approaches 0 but is not equal to 0.
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter3: The Derivative
Section3.CR: Chapter 3 Review
Problem 3CR
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Transcribed Image Text:Evaluate the limit, if it exists.
(h - 9)² - 81
h
lim
h→0
Step 1
We are asked to find the following limit.
(h - 9)² - 81
h
lim
h→0
We will first attempt to do so using direct substitution. Let f(h)
(h – 9)² – 81. We know that as f is a rational function, if 0 is in the domain of f, then the given limit is equal to f(0). The function f is not
h
is not defined at h = 0; therefore, we cannot evaluate the limit by this direct substitution.
Therefore, we will next attempt to use the quotient law, which we recall is as follows.
lim_g(x)
h→0
(h - 9)² -
h
g(x)
lim
h→0 k(x)
We note that the quotient law requires that the limit in the denominator is not equal to 0, so we must check that as follows.
lim h =
h→0
So, this means that the quotient law cannot be used
Step 2
We have observed that we cannot use direct substitution or the quotient law to find the limit. We can now attempt to simplify the given function before evaluating the limit to see if that works.
We note that if we expand the numerator and then factor it, the resulting rational expression can be simplified as follows.
h²
X h+81 - 81
- 81
lim_k(x)
h→0
6²
=h-
h
h
h
=
cannot be used
n(n-(1
h
Note that it is valid to cancel the common factor h from the numerator and denominator because h‡ 0 for the values the function is evaluated at for the limit. The variable h approaches 0 but is not equal to 0.
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