Taking derivative in different ways produce different functions

Question

Consider,

$$ f(x) = \frac{x^{n+1} - 1}{x-1}$$

If I took derivative using quotient rule,

$$ f'(x) = \frac{ (n+1)x^{n} (x-1) - (x^{n+1} -1) } { (x-1)^2}$$

Now,

$$ \lim_{x \to 1} f'(x) = \frac{n(n+1)}{2}$$

But, suppose I took derivative using product rule,

$$ f'(x) =( ( x^{n+1} -1) (x-1)^{-1})' = (n+1)x^n(x-1)^{-1} -(x-1)^{-2} (x^{n+1} -1)$$

Now, for this function if I take limit by distributing over the sum of functions, then the limit is undefined for one of them,

$$ f'(x) = \lim_{x \to 1}(n+1)x^n(x-1)^{-1} -(x-1)^{-2} (x^{n+1} -1)$$

More particularly, this limit is undefined $$ \lim_{x \to 1} \frac{ x^n}{ (x-1)}$$

So, am I missing something is the derivative of a function different based on how exactly you take the derivative of it? I mean, if both functions are actually equivalent they should have the same limits.

Edit: Why does the limit only become defined when I do algebraic manipulations? I know that when I manipulate the expression, I am indeed changing the function.

Edit: I unaccepted the answer because I saw this stack which says that for limit of a sum to exist, then the sum of limits must exist. and, the top answer says otherwise.

Answer 1

Note $$f(x) = 1 + x + x^2 + \cdots + x^n,$$ so that $$f'(x) = 1 + 2x + 3x^2 + \cdots + nx^{n-1}$$ and $$f'(1) = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}.$$

It is worth noting that although the above only applies for positive integer $n$, the result $f'(1) = n(n+1)/2$ is true for all real $n$.

Your calculation with the quotient rule is correct, but it is not clear how you evaluated the limit.

The product rule yields an equivalent expression for $f'(x)$ as in your quotient rule calculation. This is not difficult to see once you factor the term $(x-1)^{-2}$. It should be noted that evaluating the limit of each term separately is incorrect, because both terms are indeterminate. It is only when you consider them together that the limit is defined.

For instance: $$\lim_{x \to a} f(x) + g(x) \ne \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$ if the individual limits on the RHS do not exist. You can construct a trivial example with $f(x) = -g(x) = 1/x$, and $a = 0$. In your case, you are under the mistaken impression that $$\lim_{x \to 1} \frac{x^{n+1}-1}{(x-1)^2}$$ exists. I do not know why you think it does.

Answer 2

We can rearrange your second result into the first result: $$(n+1)x^n(x-1)^{-1}-(x-1)^{-2}(x^{n+1}-1)=\frac{(n+1)x^n}{x-1}-\frac{x^{n+1}-1}{(x-1)^2}=\frac{(n+1)x^n(x-1)-(x^{n+1}-1)}{(x-1)^2}$$

Answer 3

The concise answer to this is that limit is not distributive unless each limit inside the brackets is well defined.

$$ \lim_{x \to a} [ f(x) + g(x) ] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$

iff $ \lim_{x \to a} f(x)$ and $ \lim_{x \to a} g(x)$ is already well defined.