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My question is, how can I evaluate limits without L'Hopital's Rule ?

$$\lim\limits_{x \to 0} \frac{\sin(\sqrt[3]{x})}{1-\cos x}$$

Chinnapparaj R
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MAxcoder
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4 Answers4

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By standard limits we have that

$$\frac{\sin(\sqrt[3]{x})}{1-\cos x}=\frac{\sin(\sqrt[3]{x})}{\sqrt[3]{x}}\frac{x^2}{1-\cos x}\frac{\sqrt[3]{x}}{x^2}$$

and

  • $\frac{\sin(\sqrt[3]{x})}{\sqrt[3]{x}}\to 1$
  • $\frac{x^2}{1-\cos x}\to 2$

and

$$\lim_{x\to 0^+}\frac{\sqrt[3]{x}}{x^2}=+\infty$$

Note also that if we extend the existence of $\sqrt[3]{x}$ to the values $x<0$ we have

$$\lim_{x\to 0^-}\frac{\sqrt[3]{x}}{x^2}=-\infty$$

user
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For small $x$, $1-\cos x= 2\sin^2\frac{x}{2}\approx\frac{x^2}{2}$ while $\sin\sqrt[3]{x}\approx\sqrt[3]{x}$, so $\frac{\sin\sqrt[3]{x}}{1-\cos x}\approx 2x^{-5/3}$ diverges as $x\to 0$. Thus the limit is $\infty$.

J.G.
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  • I fail to see how using first-order Taylor approximation is not L'Hospital rule in a disguise. But it seems there's always this back and forth on MSE regarding this thing, so yeah, sure, why not. – Ennar Oct 18 '18 at 12:06
  • @Ennar Saying $\sin x\approx x$ isn't a Taylor-series argument; it's a geometric observation that, if you cut a very small sector from a circle, almost all of the area is in the triangle with the same vertices. – J.G. Oct 18 '18 at 12:56
  • It's not a Taylor-series argument if you prove $\lim_{x\to 0}(\sin x)/x = 1$ the way you describe. But, the notation $\approx$ leads me to think of Taylor's theorem. – Ennar Oct 18 '18 at 13:07
  • @Ennar I just take $f(x)\approx g(x)$ to mean $f/g$ has limit $1$, regardless of how we prove it. – J.G. Oct 18 '18 at 13:22
  • Ok, thanks for clarifying. – Ennar Oct 18 '18 at 13:29
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Write your term in the form $$\frac{\sin(\sqrt[3]{x})(1+\cos(x))}{\sqrt[3]{x}\frac{\sin^2(x)}{x^2}}\cdot \frac{1}{x^2}$$

and you will see that the limit doesn't exist.

Dr. Sonnhard Graubner
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You can evaluate limits without L'Hospital rule using known properties of limits:

  • $\lim_{x\to c}(f(x) + g(x)) = \lim_{x\to c}f(x)+\lim_{x\to c}g(x)$
  • $\lim_{x\to c}(f(x)g(x)) = (\lim_{x\to c}f(x))\cdot (\lim_{x\to c} g(x))$
  • $\lim_{x\to c}(f(x)/g(x)) = (\lim_{x\to c}f(x))/ (\lim_{x\to c} g(x))$

provided that all the limits exist (and denominator is not $0$ in the last one).

It can also help to use change of variables for limits:

  • If $\lim_{x\to a} g(x) = b$ and $g(x) \neq b$ on $B(a,\varepsilon)\setminus\{a\}$, then $\lim_{x\to a}f(g(x)) = \lim_{y\to b}f(y)$.

Finally, you will hardly get anywhere without using known limits such as $\lim_{x\to 0}\frac{\sin x}{x} = 1$ and similar. You must have seen some kind of list of this kind.

Good luck with your limits!

Ennar
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