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Determine the values of the two real parameters $a$ and $b$ such that

$$\lim_{x \to 0}\frac{\sin x-(ax^3+bx)}{x^3}=1$$

I immediately have thought of the Hopital theorem. We have:

$$\lim_{x \to 0}\frac{\sin x-(ax^3+bx)}{x^3}=\lim_{x \to 0}\frac{\cos x-3ax^2-b}{3x^2}=\lim_{x \to 0}\frac{1-b}{3x^2}\equiv 1$$

Now I am not able to find $a$ and $b$.

Tian Vlašić
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Sebastiano
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2 Answers2

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Hint: if the limit of a fraction tends to a finite number and the denominator tends to $0$, then also the numerator must tend to zero as if not then the limit is infinite (or doesn’t exist).

Apply this after having used l’Hôpital rule to get

$$b=1$$ and then apply l’Hôpital again.

Addendum: we want to prove that if $\lim_{x\to x_0}\frac{f(x)}{g(x)}=1$ and $\lim_{x\to x_0}g(x)=0$ then $\lim_{x\to x_0}f(x)$ is also $0$.

Indeed by definition $$\lim_{x\to x_0}\frac{f(x)}{g(x)}=1\iff \forall \varepsilon > 0\ \exists \delta |x-x_0|<\delta_1 \implies \left|\frac{f(x)}{g(x)}-1\right|<\varepsilon$$ and $$\lim_{x\to x_0}g(x)=0\iff \forall \varepsilon > 0\ \exists \delta |x-x_0|<\delta_2 \implies |g(x)|<\varepsilon$$

thus for $\delta = \min(\delta_1,\delta_2)$ \begin{align*} \left|\frac{f(x)}{g(x)}-1\right|<\varepsilon&\iff |f(x)-g(x)|<\varepsilon\cdot |g(x)|\\ &\implies |f(x)|-|g(x)|<\varepsilon|g(x)|\\ &\iff |f(x)|<(\varepsilon+1)|g(x)|\\ &\iff |f(x)|<(\varepsilon+1)|\varepsilon\\ &\implies |f(x)|<2\varepsilon \tag{$\varepsilon <1$} \end{align*} which is by definition saying that $\lim_{x\to x_0}f(x)=0$.

b00n heT
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  • +1 surely. Also I have thought your same hint. Where born the "if the limit of a fraction tends to a finite number and the denominator tends to 0, then also the numerator must tend to zero as if not then the limit is infinite (or doesn’t exist)". What is the propriety or theorem? – Sebastiano Jun 22 '23 at 08:56
  • Where I find that: So for a fraction with denominator tending to zero, for it to have finite limit, numerator should also tend to zero? – Sebastiano Jun 22 '23 at 09:01
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    I don't think this is a "theorem", but if you want to prove it formally then you have to cover the remaining cases: assume that the limit is finite but non-zero, infinite and lastly that it doesn't exist and show that in all these cases the limit can't be finite. – b00n heT Jun 22 '23 at 09:01
  • Let me provide a proof... give me 5' to write it up. – b00n heT Jun 22 '23 at 09:02
  • OK – Sebastiano Jun 22 '23 at 09:02
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    Have a look and let me know. – b00n heT Jun 22 '23 at 09:11
  • Perfect proof. I never studied such a property in my university. – Sebastiano Jun 22 '23 at 09:15
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    Glad to be of help! Cheers – b00n heT Jun 22 '23 at 09:17
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    I had thought that the numerator was zero because I was thinking of a straight line with the directorial numbers l, m, n. My geometry professor told us that if the denominator tends to zero, the numerator will also tend to zero. GRAZIEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE – Sebastiano Jun 22 '23 at 09:18
  • The intuition is correct, but then it's good to back it up using the definition as intuitions in limits can be misleading :) – b00n heT Jun 22 '23 at 09:20
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We can refer to this result

$$\lim_{x\to0}\frac{\sin x-x}{x^3}=-\frac16$$

to obtain

$$\frac{\sin x-(ax^3+bx)}{x^3}=\frac{\sin x-x}{x^3}-\frac{ax^3+(b-1)x}{x^3}$$

from which we see that we need $b=1$ and $a=-\frac 76$.

user
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