Build a table to estimate lim sin(1/x) as x approaches zero. Use multiples of pi.
I built a table, but do not understand why this limit does not exist. Please explain! Thank you!!
for a limit L to exist, you need to be able to show that
lim(h->0) |f(h)-L| -> 0
but, no matter how small an h you pick, you can show that there is a k<h such that sin(1/kπ) can be anywhere from -1 to +1
To estimate the limit as x approaches zero for the function f(x) = sin(1/x), you can create a table of values by plugging in multiples of pi for x. However, it is important to note that the limit of f(x) as x approaches zero does not exist.
Here's how you can build the table:
1) Start by choosing values of x that are multiples of pi, approaching zero. For example, you can choose x = pi/4, pi/8, pi/16, and so on.
2) Compute the corresponding values of f(x) = sin(1/x) using a calculator or by hand. For each value of x, calculate sin(1/x) based on its reciprocal.
For instance:
- When x = pi/4, the reciprocal is 4/pi, so f(pi/4) = sin(4/pi).
- When x = pi/8, the reciprocal is 8/pi, so f(pi/8) = sin(8/pi).
- Continue this process for each value of x as you approach zero.
3) Record the values of f(x) in a table, alongside their corresponding x-values.
Now, let's discuss why this limit does not exist:
As x approaches zero, the values in the table will oscillate between -1 and 1 because sin(1/x) oscillates rapidly between -1 and 1 as 1/x approaches infinity and negative infinity. This oscillation prevents the function from settling down to a specific value as x gets closer to zero.
Consider a subsequence of x approaching zero, such as (1/(2nπ)), where n is a positive integer. For this subsequence, f(x) = sin(2nπ) = 0 for every value of n. Similarly, for the subsequence (1/((2n+1)π)), f(x) = sin((2n+1)π) = 0.
Therefore, the function f(x) takes on different values (0) for different subsequences as x approaches zero. This behavior indicates that the limit of f(x) does not exist as x approaches zero.
In conclusion, although you can estimate f(x) using a table of values, you will find that the values oscillate between -1 and 1. This oscillation demonstrates that the limit of sin(1/x) as x approaches zero does not exist.