Question

Prove

$\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$

for n not belonging to rational numbers.

Answer 1

n.a^{n-1}

Answer 2

The given limit is $\lim_{x \to a} \frac{x^n - a^n}{x - a}$.

Let $f(x) = x^n$.

The expression inside the limit is $\frac{f(x) - f(a)}{x - a}$.

By the definition of the derivative, the derivative of a function $f(x)$ at a point $x=a$ is given by:

$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

provided this limit exists.

In this case, the given limit is exactly the definition of the derivative of the function $f(x) = x^n$ at $x=a$.

Therefore, the value of the limit is equal to the derivative of $x^n$ evaluated at $x=a$.

We need to find the derivative of $f(x) = x^n$ with respect to $x$. The power rule for differentiation states that $\frac{d}{dx}(x^n) = nx^{n-1}$. This rule is valid for any real number $n$.

To prove the power rule for any real number $n$, we can use logarithmic differentiation. Assume $x > 0$.

Let $y = x^n$.

Take the natural logarithm of both sides:

$\ln(y) = \ln(x^n)$

$\ln(y) = n \ln(x)$

Now, differentiate both sides with respect to $x$:

$\frac{1}{y} \frac{dy}{dx} = n \cdot \frac{1}{x}$

Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = y \cdot \frac{n}{x}$

Substitute $y = x^n$:

$\frac{dy}{dx} = x^n \cdot \frac{n}{x} = n \cdot x^{n-1}$

So, the derivative of $f(x) = x^n$ is $f'(x) = nx^{n-1}$.

This derivative is valid for any real number $n$, provided $x > 0$. Since the limit is taken as $x \to a$, we assume $a$ is in the domain where $x^n$ is defined and differentiable, which implies $a > 0$.

Now, we evaluate the derivative at $x=a$:

$f'(a) = n a^{n-1}$

By the definition of the derivative, the given limit is equal to $f'(a)$.

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

This proves the required result for any real number $n$, including non-rational numbers, assuming $a > 0$.