- #1
CuriousBanker
- 190
- 24
Hello.
Let us say that we have a graph f(x)=2x when x does not equal 3, and f(x)=undefined when x=3
The limit of f(x) as x→3, is still 6.
But what about the limit of 3+dx? Does it exist?
I am not up to the derivatives section of my class, so maybe there is something I am not understanding yet.
Because 3+dx is the smallest possible number after 3...and the limit of 3+dx=2(3+dx)=6+dx, if you just plug it into the function. But, immediately to the left of 3+dx, f(x) is not defined, so the graph isn't getting closer and closer to 3+dx as you approach it from the left hand side...because the graph is actually getting closer and closer to being undefined.
So would the limit not exist for 3+dx in this example?
Thanks
Let us say that we have a graph f(x)=2x when x does not equal 3, and f(x)=undefined when x=3
The limit of f(x) as x→3, is still 6.
But what about the limit of 3+dx? Does it exist?
I am not up to the derivatives section of my class, so maybe there is something I am not understanding yet.
Because 3+dx is the smallest possible number after 3...and the limit of 3+dx=2(3+dx)=6+dx, if you just plug it into the function. But, immediately to the left of 3+dx, f(x) is not defined, so the graph isn't getting closer and closer to 3+dx as you approach it from the left hand side...because the graph is actually getting closer and closer to being undefined.
So would the limit not exist for 3+dx in this example?
Thanks