Evaluate h(x) for the following values of x?
Consider the following function.
h(x) = (tan(x)-x)/x^3
(a) Evaluate h(x) for the following values of x. (Give your answer correct to six decimal places.)
x =
a) 1
b) 0.5
c) 0.1
d) 0.05
e) 0.01
f) 0.005
what is f(x)
(b) Guess the value of the limit of h(x) as x approaches 0. (If it does not exist, enter NONE.)
We want to "guess" that limh(x)=1/3 if x->0
h(1)=(tan(1)-1)/1^3=0.557408
h(0.1)=0.334672
h(0.005)=(tan(0.005)-0.005)/0.005^3=0.333342
To evaluate h(x) for the given values of x and find the value of the limit of h(x) as x approaches 0, we will substitute each x into the expression for h(x) and perform the necessary calculations. Let's go through each value of x one by one.
(a) To evaluate h(x) for x = 1:
Substituting x = 1 into the expression for h(x):
h(1) = (tan(1) - 1)/1^3 = (approximately 0.557407 - 1)/1 = approximately -0.442593
Similarly, substituting the other values of x into the expression for h(x), we get:
(b) For x = 0.5:
h(0.5) = (tan(0.5) - 0.5)/0.5^3 = (approximately 0.546302 - 0.5)/0.125 = approximately 0.370408
(c) For x = 0.1:
h(0.1) = (tan(0.1) - 0.1)/0.1^3 = (approximately 0.100334 - 0.1)/0.001 = approximately 0.233400
(d) For x = 0.05:
h(0.05) = (tan(0.05) - 0.05)/0.05^3 = (approximately 0.050010 - 0.05)/0.000125 = approximately 0.003200
(e) For x = 0.01:
h(0.01) = (tan(0.01) - 0.01)/0.01^3 = (approximately 0.010001 - 0.01)/0.000001 = approximately 0.001000
(f) For x = 0.005:
h(0.005) = (tan(0.005) - 0.005)/0.005^3 = (approximately 0.005001 - 0.005)/0.000000125 = approximately 0.000800
Now, let's consider the value of the limit of h(x) as x approaches 0.
(b) To guess the value of the limit of h(x) as x approaches 0, we can substitute smaller and smaller values of x into the expression for h(x):
As x approaches 0, the values of h(x) become closer to -0.442593, which we obtained for x = 1. Therefore, we can guess that the limit of h(x) as x approaches 0 is approximately -0.442593.
Note that this is only a guess based on the calculations we did for x-values approaching 0, and it does not prove the actual limit value. To find the exact limit, more advanced mathematical tools and techniques are required.