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$$\lim_{x \rightarrow 0} \frac{{(1-x)}^{k}-({1-k x})}{x^2}$$ $( k>0)$

2)

$$\lim_{x \rightarrow z} \frac{x^m-z^m}{x^n-z^n}$$ $(z>0, m, n \in \mathbb{R}, n \neq 0)$

What I’ve done for: for 1)

I have applied l’Hospital, then I have $\lim_{x \rightarrow 0} \frac{k (1-x)^{k-1} (-1)+k}{2x} = \lim_{x \rightarrow 0} \frac{k ((x-1)^{k-1}+1)}{2x}$ but I don’t know how to continue from here. If $x$ converges to zero the nominator also converges to zero, right?

And for the second one, I have also applied l’Hospital

$$\lim_{x \rightarrow z} \frac{x^m-z^m}{x^n-z^n}= \lim_{x \rightarrow z} \frac{mx^{m-1} - mz^{m-1}}{nx^{n-1}-nz^{n-1}}$$ but it doesn’t look like that this would help me?

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  • $\begingroup$ Why $(1-x)^{k-1}(-1)=(x-1)^{k-1}$? It holds only for even $k$. $\endgroup$
    – richrow
    Jun 6, 2021 at 12:06
  • $\begingroup$ In the second problem you differentiate with respect to $x$, so $(z^m)'=0$ and $(z^n)'=0$. $\endgroup$
    – richrow
    Jun 6, 2021 at 12:07
  • $\begingroup$ @richrow: It is not mentioned in the OP whether $k4 is even or not. In any case, L'Hospital is applied twice here. One can also fins other ways to get the limit without it. $\endgroup$
    – Oliver Díaz
    Jun 6, 2021 at 12:10
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    $\begingroup$ 2. If $z\neq0$ or $m>n$, then as $x\rightarrow z$ $$ \frac{x^m-z^m}{x^n-z^n}\sim\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}x^{m-n}\xrightarrow{x\rightarrow z}\frac{m}{n}z^{m-n}$$ $\endgroup$
    – Oliver Díaz
    Jun 6, 2021 at 12:13
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    $\begingroup$ If $k=1$, your limit is $0$. If $k\neq 1$, then as $x\rightarrow0$, $$ \frac{{(1-x)}^{k}-({1-k x})}{x^2}\sim \frac{-k(1-x)^{k-1}+k}{2x}\sim \frac{k(k-1)(1-x)^{k-2}}{2}\xrightarrow{x\rightarrow0}\frac{k(k-1)}{2}$$ $\endgroup$
    – Oliver Díaz
    Jun 6, 2021 at 12:17

2 Answers 2

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Lemma: $\displaystyle\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$

Proof: Use the fact that $x^n-a^n = (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\dots+a^{n-2}x+a^{n-1})$, if $n$ is a natural number. Otherwise use L'Hopital rule.

Now our first limit after your use of L'Hopital rule becomes \begin{align} \lim_{x \rightarrow 0} \frac{{(1-x)}^{k}-({1-k x})}{x^2} &= \frac k2\lim_{1-x\to1}\frac{1-(1-x)^{k-1}}{1-(1-x)}\\ &=\frac k2\cdot (k-1)1^{k-2} \\ &= \frac{k(k-1)}{2} \end{align}

Second limit immediately becomes $$\lim_{x\to z}\dfrac{\dfrac{x^m-z^m}{x-z}}{\dfrac{x^n-z^n}{x-z}}=\frac mnz^{m-n}$$

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  • $\begingroup$ Thank you so much, the only thing what I don’t entirely understand is why did you change the limits from $ x \rightarrow 0$ to $ 1 - x \rightarrow 1$? Which added benefit does this have? $\endgroup$
    – Parinn
    Jun 6, 2021 at 12:28
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    $\begingroup$ @Parinn, To get into appropriate form for the lemma to be applicable. $\endgroup$
    – Martund
    Jun 6, 2021 at 12:29
  • $\begingroup$ @Martund: If $f(x)=x^n$, then by definition $$f'(a)=\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\lim_{x \to a}\frac{x^n-x^a}{x-a} \, .$$So I don't like using L'Hôpital's rule here. $\endgroup$
    – Joe
    Jun 6, 2021 at 13:19
  • $\begingroup$ @Joe, Yeah, this is the definition of derivative. But we evaluate that derivative using $x^n = e^{n\log x}$ and then differentiating using chain rule, in practice. So differentiating like that and then equating to definition of derivative gives us this limit. $\endgroup$
    – Martund
    Jun 6, 2021 at 14:46
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Your assertion that if denominator tends to $0$, so should numerator, is in general, baseless, although under certain conditions it holds. Moreover, you have incorrectly assumed that $(1-x)^k(-1)=(x-1)^k$. This holds only if: $$(1-x)^k(-1)=(x-1)^k(-1)^{k+1}=(x-1)^k$$ Which means that $k$ is odd. However, if you proceed with $(1-x)^k(-1)$, you will see that $\frac 00$ indeterminate form is retained and hence L'Hôpital rule can be applied again.

Alternative:

Instead of applying the L'Hôpital rule, it would be more useful to take advantage of Newton's expansion: $$(1-x)^k=1-kx+\frac {k(k-1) x^2}{2}-...$$ This is an infinite series, but writing three terms suffices for our purpose.

For the second one, you have not correctly applied the rule. The differentiation is to be done with respect to $x$, and $z$ is to be simply treated as a constant. I believe you can do it now, the value of the limit would depend upon which of $m$ and $n$ is greater.

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    $\begingroup$ No it would not. You can perfectly use L'Hospital here as (1) the functions involve are differentiable, (2) you get indeterminancines of the form $0/0$ (3) the limit of the $\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$ exists in both cases. $\endgroup$
    – Oliver Díaz
    Jun 6, 2021 at 12:13
  • $\begingroup$ I do not claim that L'Hôpital cannot be applied. I merely suggested an alternate approach, since OP was confused. I also pointed out the error in his work, as you'll notice $\endgroup$
    – Ritam_Dasgupta
    Jun 6, 2021 at 12:17
  • $\begingroup$ No, read the claim OP makes. He says that since denominator tends to $0$, so should numerator. That is most definitely incorrect $\endgroup$
    – Ritam_Dasgupta
    Jun 6, 2021 at 12:20
  • $\begingroup$ And there is no need for $k$ to be an integer in the generalized binomial theorem (also called Newton's series) which I have applied. $\endgroup$
    – Ritam_Dasgupta
    Jun 6, 2021 at 12:20
  • $\begingroup$ This is actually an extremely elementary technique to solve limits. You can derive the same series using Taylor polynomial theory. $\endgroup$
    – Ritam_Dasgupta
    Jun 6, 2021 at 12:46

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