Find $\lim_{x \to 1} \frac{x^{\pi} - x^e}{x-1}$

Question

I have trouble calculating this limit algebraically (without L'Hospital's rule):

$$\lim_{x \to 1} \frac{x^{\pi} - x^e}{x-1}$$

Substituting 1 gives indeterminate form. If this was e.g. $(x^5 - x^2)$ in the numerator, one could easily factor out the $(x-1)$, because $(x^5 - x^2) = x^2 (x - 1) (x^2 + x + 1)$. But I don't know what to do in my problem, because the exponents are not integers.

EDIT: The problem is from a calculus textbook and is before derivatives are introduced so I assume it can be found without derivatives.

Answer 1

Hint: write $$ \lim_{x\to 1}\frac{x^{\pi}-x^e}{x-1}=\lim_{x\to 1}\frac{x^{\pi}-1}{x-1}-\lim_{x\to 1}\frac{x^e-1}{x-1}$$ and then note that each term is a derivative.

Answer 2

Try

$$\lim_{x\to 1}x^{e}\frac{x^{\pi-e}-1}{x-1}$$

Now it depends on what you are "allowed" to use. If you assume binomial expansion works for real exponends, then that's it (this is the Taylor expansion, or the well-known first order approximation $(1+\epsilon)^n\approx 1+n\epsilon$).

Answer 3

Ok so for integer $n$, its clear $$\frac{x^n-1}{x-1}\to n$$ for a fraction $\frac{p}{q}$ let $x=y^q$ and then $$\frac{x^{\frac{p}{q}}-1}{x-1}=\frac{y^p-1}{y^q-1}\to \frac{p}{q}$$
Now for any $a$ we have

$$\lim\limits_{x\to 1^+}\frac{x^a-1}{x-1}=\lim\limits_{x\to 1^+}\sup\{\frac{x^r-1}{x-1}|r\in \mathbb{Q}, r<a\}$$ $$=\sup\{\lim\limits_{x\to 1^+}\frac{x^r-1}{x-1}|r\in \mathbb{Q}, r<a\}=a$$ thus your limit is $$\pi-e$$