# Algebra of limits vs. L'Hospital

We have to evaluate the following limit:

$$\displaystyle\lim_{x\to 0} \dfrac{x - \sin(x)}{x^3}$$

When I evaluate it using L'Hospital rule I get $\dfrac16$ but when I simplify the limit using simple algebraic rules of limit I get $0$.

We can use the this rule: $\displaystyle\lim [f(x) \pm g(x)] = \lim[f(x)] \pm \lim[g(x)]$ to simplify the given limit.

$$\lim_{x\to 0} \dfrac{x - \sin(x)}{x^3} = \lim_{x\to 0}\dfrac{x}{x^3} - \lim_{x\to 0}\dfrac{\sin(x)}{x^3}$$

Also, $\lim[f(x).g(x)] = \lim[f(x)].\lim[g(x)]$.

$$\therefore \lim_{x\to 0}\dfrac{x}{x^3} - \lim_{x\to 0}\dfrac{\sin(x)}{x^3} = \lim_{x \to 0}\dfrac{1}{x^2} - \left(\lim_{x\to 0} \dfrac{\sin(x)}{x} \right)\left(\lim_{x\to 0} \dfrac{1}{x^2}\right)$$

Now, since, $\lim_{x\to 0} \dfrac{\sin(x)}{x} = 1$,

$$\lim_{x \to 0}\dfrac{1}{x^2} - \left(\lim_{x\to 0} \dfrac{\sin(x)}{x} \right)\left(\lim_{x\to 0} \dfrac{1}{x^2}\right) = \lim_{x \to 0}\dfrac{1}{x^2} - \lim_{x\to 0} \dfrac{1}{x^2} = \lim_{x \to 0}\left(\dfrac{1}{x^2} - \dfrac{1}{x^2}\right) = 0$$

Why am I getting $0$ when I use algebra of limits.

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$\infty - \infty \neq 0$ –  Tim Seguine May 25 '14 at 9:51
@TimSeguine I fail to see how your comment is relevant to my answer. –  ShuklaSannidhya Aug 3 '14 at 6:41
It is relevant because that is precisely what you tried to do. Look at your second to last step before the =0 you have two infinite limits with a minus sign. Everything after that is pointless because "you done screwed up" –  Tim Seguine Aug 6 '14 at 10:34

Rules like

$$\lim[f(x) \pm g(x)]=\lim[f(x)] \pm \lim[g(x)]$$

are only true if all the limits involved converge to finite numbers. Almost all the limits involved in your argument turn out to approach infinity.

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The problem is that $\lim_{x\to 0}\dfrac{\sin(x)}{x^3} = \infty$ so you cannot evaluate the limits independently.

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There are two different kinds of mistakes here:

1) the algebra of limits assumes that the limits do exist. Here after the split none of $x/x^{3}$ and $\sin x/x^{3}$ tend to a limit. This has been communicated in other answers also.

2) the second mistake is bit tricky to figure out. In the last step the calculation is like $$L = \lim_{x \to a}f(x) - \lim_{x \to a}g(x)\lim_{x \to a}h(x)$$ where $\lim_{x \to a}g(x) = 1$ and then this limit $1$ is substituted to get $$L = \lim_{x \to a}f(x) - \lim_{x \to a}h(x)$$ and in this case $f(x) = h(x)$ and the next step is $$L = \lim_{x \to a}f(x) - h(x) = 0$$ This part again violates the rule of algebra of limits mentioned in part 1) above.

But my concern is the substitution of the $\lim_{x \to a}g(x)$ by $1$ during calculation of the limit $L$. There are only two scenarios where such a substitution can be made.

Let $\lim_{x \to a}g(x) = A$ exist then

• $\lim_{x \to a}\{f(x) \pm g(x)\} = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x) = \lim_{x \to a}f(x) \pm A$ irrespective of whether $\lim_{x \to a}f(x)$ exists or not.
• $\lim_{x \to a}f(x)\cdot g(x) = \lim_{x \to a}f(x)\cdot\lim_{x \to a}g(x) = A\lim_{x \to a}f(x)$ provided $A \neq 0$ irrespective of whether $\lim_{x \to a}f(x)$ exists or not.

During calculation of limit of a complicated expression we normally apply rules of algebra of limits and thereby the whole limit is expressed as an algebraic combination of limits of smaller expressions. We can replace one of the smaller limit expression by their limiting value only in the two cases mentioned above. This is explained in more depth in my blog post under the heading "misuse of rules of limits".

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Consider

$$\lim_{x\to 0} f(x)g(x) = \lim_{x\to 0} x\dfrac{f(x)g(x)}{x} = \left(\lim_{x\to 0} x\right)\left(\lim_{x\to 0} \dfrac{f(x)g(x)}{x}\right) = 0\times \left(\lim_{x\to 0}\dfrac{f(x)g(x)}{x}\right)= 0$$

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In sum, l'Hopital's rule can be used here, but you cannot take the limit of $x/x^3$ and $(\sin x)/x^3$ separately and then take the difference because both limits are $\infty$ as $x\downarrow 0$.

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