True or False. Explain.
If lim f(x)=L
x → a
then
f(a)=L
sure looks like it to me
Not at all. Just because the limit exists, that does not mean the function at x=a has the value L. It might not even be defined at all.
recall your previous example, f(x)=(x^2+x-2)/(x^2-1), where the limit as x->1 was 1.5, but the function was not defined there.
There was a hole. In fact, the function might have been defined to be 10 there, so while the limit exists, the actual value of the function could be anything or undefined.
Evaluating the limit is not the same as evaluating the function!
The statement is false. Let me explain why.
The limit of a function as x approaches a particular value, denoted as lim f(x) = L as x approaches a, does not necessarily imply that f(a) equals L.
The concept of a limit allows us to understand the behavior of a function near a specific point without actually evaluating the function at that point. It provides information about the behavior of the function as x approaches the given value, but it doesn't guarantee that the value of the function at that point will be the same as the limit.
In other words, even if the limit exists and is equal to a certain value L, it does not automatically imply that the function is defined or takes the value L at that exact point.
To determine whether f(a) equals L, we need to evaluate the function directly at a, not just take the limit. In some cases, the value of the function at a may equal L, but in other cases, it may not.
For example, consider the function f(x) = x^2. If we take the limit of f(x) as x approaches 0, we have lim f(x) = 0 as x approaches 0. However, f(0) = 0, which is not equal to the limit. This demonstrates that the statement "f(a) equals L when lim f(x) = L as x approaches a" is not always true.
In summary, the truth value of the statement "If lim f(x) = L as x approaches a, then f(a) = L" is false. The limit of a function does not guarantee that the function takes the same value as the limit at the specific point a.