Are the following two limits the same?

Question

If we assume that the $\lim_{x\to\infty} f(x)$ exists (let's call it L). Then is the $\lim_{x\to\infty} f(x+1)$ also equal L? Where $f(x)$ is within the domain of all positive integers.

Firstly, I have tried to solve this problem by looking for a proof by contradiction however every function I have tried seems to produce the same limit. I am not sure how to go about proving this statement to be true with a rigorous proof. Any help or hints as to where I should start would be very much appreciated.

Answer 1

The limits are the same.

From the definition, $\lim_{x \to \infty} f(x) = L$ means that, for every $\epsilon > 0$ there is a $v =v(\epsilon)$ such that $x > v \implies |f(x)-L| < \epsilon$.

To do this for $f(x+1)$ just involves changing the result of $v(\epsilon)$ by $1$.

Answer 2

Letting $x$ go to infinity implies that we let the new argument, $(x+1)$, go to infinity as well, and so the limit of $f(x+1)$ is again (by assumption) equal to L.

Answer 3

The standard definition gives the standard proof, as marty cohen has shown. It says for limit to infinity that given any positive error margin there is a point beyond which the function stays within the error margin around the limit value. However, this is somewhat backward to the intuition, which is that as the input value goes to infinity the output value goes to a limit. So it is often useful to think of limits in the intuitive sense as follows:

If $f(x) \to L$ as $x \to \infty$:

  As $x \to \infty$:

    We also have $x+1 \to \infty$.

    Thus $f(x+1) \to \infty$.