# Limit with $\tan$ L'Hopital

I have kind of a simple and maybe stupid question but $$\lim_{x\to 0} \frac{(\tan x-x)}{x^3}$$ why am I not allowed to split the limit like this : $$\lim_{x\to 0} \frac {\tan x}{x^3} -\frac{x}{x^3}$$ which equals $0$? I came up with the right answer after using L'Hopital.

• Why is it equal to 0? Commented May 1, 2017 at 15:09
• Well when I tried to split it it would be something like $$\lim_{x\to 0} \frac {1}{x^2} -\frac{1}{x^2}$$ which equals 0 using that $$\lim_{x\to 0} \frac {tg x}{x} =1$$
– Lola
Commented May 1, 2017 at 15:13
• My bad I calculated for x going to inf
– Lola
Commented May 1, 2017 at 15:14
• It yields to 1/3 Commented May 1, 2017 at 15:19
• Yes that's the answer I found too after using L'H
– Lola
Commented May 1, 2017 at 15:19

$\infty - \infty$ does not a zero make, when it comes to limits.

For example, the limit \begin{equation*} \lim_{x \rightarrow \infty} \frac{x^5 - x^4}{x} \end{equation*} is clearly $\infty$, but if you tried to "split" you would get $\infty - \infty$. The problem is that the two quantities go to $\infty$ at a different rate (here, the positive term is much faster than the negative term).

• Yes I know, I forgot that x goes to 0
– Lola
Commented May 1, 2017 at 15:15
• It works the same way, for instance, I could have $x \rightarrow 0$ in my example and then replace $x^5$ with $x^{-5}$. Point being, you can't simplify $\infty - \infty$ to $0$. Commented May 1, 2017 at 15:20

$$\lim_{x\to 0} \frac {\tan x}{x^3} -\frac{x}{x^3}=\lim_{x\to 0} \frac {\tan x}{x}\cdot\frac{1}{x^2} -\frac{1}{x^2}$$

and once,

$$\lim_{x\to 0} \frac {\tan x}{x}=1$$

your limit is $$1\cdot\infty - \infty=\infty - \infty$$ which is not $0$.

There are two mistakes depending on how you are trying to solve it. If you write $$\lim_{x \to 0}\frac{\tan x - x}{x^{3}} = \lim_{x \to 0}\frac{\tan x}{x^{3}} - \lim_{x \to 0}\frac{1}{x^{2}}\tag{1}$$ then you are violating the laws of algebra of limits. The above step is valid only when at least one of the limits on the right exists (and is finite).

On the other hand if you write $$\lim_{x \to 0}\frac{\tan x - x}{x^{3}} = \lim_{x \to 0}\frac{\tan x}{x^{3}} - \frac{1}{x^{2}}\tag{2}$$ then this step is fine. There is no split of limits. Only the expression has been manipulated according to the laws of algebra. Next we can write this as $$\lim_{x \to 0}\frac{\tan x}{x}\cdot\frac{1}{x^{2}} - \frac{1}{x^{2}}\tag{3}$$ and there is no problem till this point. The mistake is committed when we replace the sub-expression $(\tan x)/x$ with its limit $1$ to get $$\lim_{x \to 0}\frac{1}{x^{2}} - \frac{1}{x^{2}} = 0\tag{4}$$ This is not allowed via any rules of limits. Please see this answer for more details on when a sub-expression can be replaced by its limit while evaluating the limit of a complicated expression.

Also note that there is no issue like $\infty - \infty$ here as other users are indicating. The equation $(4)$ is correct. What is incorrect is the transition from equation $(3)$ to equation $(4)$.

One can only apply the rule $$\lim_{x\to a}(f(x) + g(x)) = \lim_{x\to a}f(x) + \lim_{x\to a} g(x)$$

if the expression on the right is defined. In your case, it gives $+\infty - (+\infty)$ which is in indeterminate form (not defined). Therefore, you should try another method than splitting.