how can I prove the following: Prove that $$ \lim_{x \to y} \frac{x^k-y^k}{x-y} = ky^{k-1} $$ for a fixed $y\in\mathbb R$ and $k\in\mathbb N$. I already tried a lot but somehow I don't come to the correct conclusion
Asked
Active
Viewed 69 times
1
-
Did you tried to use L'Hopital rule? – rafa11111 Dec 17 '17 at 13:24
-
Please see https://math.meta.stackexchange.com/questions/5020/ – Angina Seng Dec 17 '17 at 13:24
-
1@rafa11111 or just use the definition of derivative. – Angina Seng Dec 17 '17 at 13:25
-
We never heard about the L'Hospital or using the definition of derivative in lecture. Are there other methods? – maa Dec 17 '17 at 13:29
-
I'm afraid that no. This limit is of the kind that leads to something like $0/0$ with no simple way of simplifying it. Are you familiar to the concept of the derivative? – rafa11111 Dec 17 '17 at 13:38
-
I learned it in school but not at university (yet), but go on – maa Dec 17 '17 at 13:40
2 Answers
1
Using $$x^k -y^k= (x-y)\left(\sum_{i=0}^{k-1} x^{k-i-1}y^i\right)$$
the limit is $$\lim_{x \to y}\sum_{i=0}^{k-1} x^{k-i-1}y^i = \sum_{i=0}^{k-1} y^{k-1}=ky^{k-1}$$
user8277998
- 2,666
0
The derivative of a function $f(x)$ is $$ \frac{df}{dx} = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $$
Let $f(x) = x^k$ and $h = x-y$. Then: $$ \frac{df}{dx} = \lim_{h \to 0} \frac{(x+h)^k-x^k}{h} $$
From the binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem) you can see that $$ \frac{df}{dx} = \lim_{h \to 0} \frac{(x^k + k x^{k-1} h + \cdots )-x^k}{h} = \lim_{h \to 0} \frac{k x^{k-1} h + \cdots}{h} $$ The terms included in the $\cdots$ involve higher powers of $h$, i.e., $h^2$, $h^3$ and so on. If you divide all by $h$, you will have $$ \frac{df}{dx} = \lim_{h \to 0} k x^{k-1} + \cdots $$ If you make $h\to 0$, the terms included in $\cdots$ vanishes, and you have $$\frac{df}{dx} = k x^{k-1}$$ Then: $$ \lim_{x \to y} \frac{x^k-y^k}{x-y} = k x^{k-1} $$
Naturally, you can see that $df/dx = k x^{k-1}$ using derivative rules.
rafa11111
- 1,514
- 2
- 10
- 19