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I know how to evaluate $\underset{x \to 1}{\lim} \frac{x^{2019} - 1}{x-1}$ using the definition of the derivative and the power rule. If $f(x) = x^{2019}$, then $$\lim_{x \to 1} \frac{x^{2019} - 1}{x-1} = \lim_{x \to 1} \frac{x^{2019} - 1^{2019}}{x-1} = f'(1) = 2019(1)^{2018} = 2019.$$ I assume the problem from the title is similar since the answer is $144$ according to Wolfram Alpha, but I can't wrap my head around it.

Edit: I want to know how to do this with the definition of the derivative, not L'Hospital's.

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    $\begingroup$ Alas, the others have beaten me to all of my good answers, but welcome, Michael. Excellent first question. $\endgroup$ – The Count May 2 at 22:20
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$$\lim_{x \to 1} \dfrac{x^{2019}-x^{1875}}{x-1} = \lim_{x \to 1} \left( x^{1875}\cdot \dfrac{x^{144}-1}{x-1}\right)$$

The product of the limits is equal to the limit of the product so long as both limits exist.

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  • $\begingroup$ Nice, I didn't think about it :) $\endgroup$ – Botond May 2 at 22:01
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Hint: $$x^{2019}-x^{1875}=(x^{2019}-1)-(x^{1875}-1)$$

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A solution that just uses the definition of the derivative only. Specifically, consider $$f(x) = x^{a}-x^{b}.$$ Then $$f'(1) = \lim_{x \to 1} \frac{f(x)-f(1)}{x-1} = \lim_{x \to 1}\frac{x^a-x^b}{x-1}.$$ Could you prove $$f'(1) = a-b?$$

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More generally, if $\sum_{k=1}^n a_k = 0 $ then $\lim_{x \to 1} \dfrac{\sum_{k=1}^n a_k x^{b_k}}{x-1} =\sum_{k=1}^n a_kb_k $.

Two easy proofs.

1) L'Hôpital:

$\lim_{x \to 1} \dfrac{\sum_{k=1}^n a_k x^{b_k}}{x-1} =\lim_{x \to 1} \dfrac{\sum_{k=1}^n a_kb_k x^{b_k-1}}{1} =\sum_{k=1}^n a_kb_k $.

2) Elementary with positive integer exponents:

Since, if $b$ is a positive integer, $\lim_{x \to 1} \dfrac{x^b-1}{x-1} =\lim_{x \to 1} \sum_{j=0}^{b-1} x^j =b $,

$\begin{array}\\ \lim_{x \to 1} \dfrac{\sum_{k=1}^n a_k x^{b_k}}{x-1} &=\lim_{x \to 1} \dfrac{\sum_{k=1}^n a_k (x^{b_k}-1+1)}{x-1}\\ &=\lim_{x \to 1} \left(\dfrac{\sum_{k=1}^n a_k (x^{b_k}-1+1)}{x-1}\right)\\ &=\lim_{x \to 1} \left(\dfrac{\sum_{k=1}^n a_k (x^{b_k}-1)}{x-1}+\dfrac{\sum_{k=1}^n a_k }{x-1}\right)\\ &=\lim_{x \to 1} \dfrac{\sum_{k=1}^n a_k (x^{b_k}-1)}{x-1}\\ &=\lim_{x \to 1} \sum_{k=1}^n \dfrac{a_k (x^{b_k}-1)}{x-1}\\ &=\lim_{x \to 1} \sum_{k=1}^n a_k\dfrac{ x^{b_k}-1}{x-1}\\ &= \sum_{k=1}^n a_k\lim_{x \to 1}\dfrac{ x^{b_k}-1}{x-1} \qquad\text{since all the limits exist}\\ &= \sum_{k=1}^n a_kb_k \qquad\text{if all } b_k \text{ are positive integers}\\ \end{array} $

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$$(x-1)\cdot\sum_{k=0}^{n-1} x^k = x^n - 1\ \ \implies \ \ \lim_{x\to 1}\dfrac{x^n - 1}{x-1} = n$$

Apply to

$$\dfrac{x^{n+d} - x^d}{x-1}=x^d\cdot\dfrac{x^n - 1}{x-1}$$

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$$x^a-x^b=(x^a-1)-(x^b-1)=(x-1)(x^{a-1}+x^{a-2}+\cdots1)-(x-1)(x^{b-1}+x^{b-2}+\cdots1).$$

Simplifying by $x-1$ and evaluating at $1$,

$$a-b.$$

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Edit: I want to know how to do this with the definition of the derivative, not L'Hospital's.

It is worth pointing out that this distinction does not make much sense.

When the denominator is $x-a$, what L'Hospital's rule would get you is neither more nor less than "hey, look, this is actually the definition of a derivative!"

In general, you want to find $$ \lim_{x\to a}\frac{g(x)}{x-a}$$ for some function $g$ with $g(a)=0$. In your concrete example you have $a=1$ and $g(x)=x^{2019}-x^{1875}$.

Applying L'Hospital naively would then give you $\frac{g'(x)}{1}$. But if you want to avoid L'Hospital and "instead" use the definition of a derivative, you can do that too. Just note that $g(a)=0$, so $$ \lim_{x\to a}\frac{g(x)}{x-a} = \lim_{x\to a}\frac{g(x)-0}{x-a} = \lim_{x\to a}\frac{g(x)-g(a)}{x-a} $$ and the right-hand side of this is exactly* the definition of $g'(a)$.

In either case you would now proceed to find the derivative of $x^{2019}-x^{1875}$ symbolically.

* Perhaps, depending on which textbook you get your definition from, you need to switch ariables to $h=x-a$ to get $ \lim\limits_{h\to 0}\frac{g(a+h)-g(a)}{h}$ instead.


Small caveat: A more careful formulation of L'Hospital's rule would give you $\lim_{x\to 0}\frac{g'(x)}{1}$ rather than $\frac{g'(x)}{1}$ itself. But since a derivative cannot have jump discontinuities (due to Darboux's theorem), if this limit and $g'(a)$ both exist, they must be equal. So this version of L'Hospital gives you less than "hey look, this is actually the definition of a derivative", if the derivative of $g$ happens to be discontinuous at $a$.

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