How do we evaluate: $\lim_{x \to y}{\frac{x^n - y^n}{x - y}} = ny^{n - 1}$

Question

So I was doing problems from Michael Spivak's calculus book & I've faced this question $$\lim_{x \to y}{\frac{x^n- y^n}{x - y}}$$

The answer guide isn't clear/detailed. It's saying that the answer is $ny^{n-1}$ can someone tell me why is that the correct answer? :(

Answer 1

I've just came out with the solution guys some friend of mine explained it to me and sent me a video it's actually easy but it uses the Binomial Theorem in the steps (I mean, if you wish to do it without derivatives, which what I wanted in the first place) So first of all, you assume that you wanna approach y from the positive side, and the limit of y from the positive side is just the same as (y + a) where a is a small number that tends to zero a → 0 The limit of the function should be the same from both sides, and we wish to rewrite it as the following $$\lim_{x \to y^+}{\frac{x^n- y^n}{x - y}} = \lim_{x \to y}{\frac{x^n- y^n}{x - y}}$$

And lim_x→y⁺ X is just y+a, where: a → 0

then

$$\lim_{x \to y^+}{\frac{x^n- y^n}{x - y}} = \lim_{a \to 0}{\frac{(y + a)^n- y^n}{(y + a) - y}}$$

then we'll take y as a common factor from the first term in the numerator

$$\lim_{a \to 0}{\frac{[y(1 + {\frac{a}{y}})]^n- y^n}{(y + a) - y}}$$ and the 2 terms in denominator: -y and y, will cancel out each other, and we can also distribute the power "n" for the first term $$\lim_{a \to 0}{\frac{[y^n(1 + {\frac{a}{y}})^n- y^n}{a}}$$

and now we're having 2 terms subtracted from each other in the numerator, they both are having the variable yⁿ, so we can take yⁿ as a common factor

$$\lim_{a \to 0}{\frac{y^n[(1 + {\frac{a}{y}})^n- 1]}{a}}$$

Now I KNOW that may seem hard or complex or even meaningless but trust me after those steps everything becomes easier Right now we need to use thee Binomial Theorem, this might be a little bit complicated to understand in text but if you take a paper and a pen & give it a try you'll find that the next step isn't hard the term (1 + a/y)ⁿ can be expanded using the binomial theorem, the first term will be +1, the second will be n × a/y, the third will be ⁿC₂ × (a/y)², and then you repeat the process n times, and (a/y)ⁿ will be our last term, we will have a -1 at the end from our original expression though, so after the expansion it will look like 1+(a/y)+ⁿC₂ × (a/y)ⁿ+...-1

ⁿC₂ itself could be simplified

$$^nC_2 = {\frac{n!}{(n-2)!2!}} = {\frac{n(n-1)(n-2)!}{(n-2)!2!}}$$ the term (n-2)! in the denominator will cancel out the one in the numerator $${\frac{n(n-1)}{2!}}$$

Now, our main expression will be:

$$\lim_{a \to 0}{\frac{y^n[1 + n{\frac{a}{y}}+{\frac{n(n-1)}{2!}}({\frac{a}{y}})^2+...+({\frac{a}{y}})^n- 1]}{a}}$$

And now, the 1 and the -1 will cancel out each other, wo we will have :

$$\lim_{a \to 0}{\frac{y^n[n{\frac{a}{y}}+{\frac{n(n-1)}{2!}}({\frac{a^2}{y^2}})+...+({\frac{a^n}{y^n}})]}{a}}$$

Now if you pay more attention, you'll notice that a is a common factor in all the terms of the expansion, swap it out!

$$\lim_{a \to 0}{\frac{y^na[{\frac{n}{y}}+{\frac{n(n-1)}{2!}}({\frac{a}{y^2}})+...+({\frac{a^{n-1}}{y^{n-1}}})]}{a}}$$

Now we have an a on the numerator and on the denominator, they'll cancel out each other we'll have $$\lim_{a \to 0}{y^n[{\frac{n}{y}}+{\frac{n(n-1)}{2!}}({\frac{a}{y^2}})+...+({\frac{a^{n-1}}{y^n}})]}$$

Now, substitute for all the a's in the expression with 0, and you'll have all the terms containing a canceled out you'll have the following

$$y^n * {\frac{n}{y}} = ny^{n-1}$$

so, here we came out with the famous rule: $$\lim_{x \to y^+}{\frac{x^n- y^n}{x - y}} = ny^{n-1} $$

and that's all :)

Answer 2

That limit is precisely the limit definition of the derivative of $x^n$ at $x = y$. Recall $$ f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}. $$ Consider this when $f(x) = x^n$ and $a = y$.

But maybe this question is being asked in Spivak's book before derivatives have been introduced. (EDIT: It is; see below.) You didn't tell us where it appears in that book. If it appears before the chapter on derivatives, use the standard factorization of $x^n - y^n$ with a factor of $x-y$: $$ x^2-y^2 = (x-y)(x+y), $$ $$ x^3 - y^3 = (x-y)(x^2+xy+y^2), $$ $$ x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3) $$ and so on (the factorization for general $n$ has already been mentioned in a comment above).

EDIT: I found the problem in Spivak's book: part (iv) of problem 1 in Chapter 5. Here is the whole exercise.

Find the following limits. (These limits all follows, after some algebraic manipulations, from the various parts of Theorem 2; be sure you know which ones are used in each case, but don't bother listing them.)

(i) $\lim_{x \to 1} (x^2-1)/(x+1)$.

(ii) $\lim_{x \to 2} (x^3-8)/(x-2)$.

(iii) $\lim_{x \to 3} (x^3-8)/(x-2)$.

(iv) $\lim_{x \to y} (x^n-y^n)/(x-y)$.

(v) $\lim_{h \to 0} (\sqrt{a+h}-\sqrt{a})/h$.

That phrase "some algebraic manipulations" in the instructions show that $n$ is meant to be an integer, and (EDIT) Theorem $2$ in Chapter $5$ of the book is about limits being preserved under adding, multiplying, and reciprocating (for a nonzero limit). In fact I'd say $n$ is meant to be a positive integer, although the limit formula is true if $n$ is an arbitrary integer, or even an arbitrary real number, but to handle real exponents goes beyond mere algebraic manipulations. The reader of Spivak is supposed to figure out how to factor $x^n - y^n$ by having already done this in earlier parts for $x^2-1$ and $x^3-8$. OP, were you able to do parts (i), (ii), and (iii) already?