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? – CY Aries May 1 '17 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 May 1 '17 at 15:13
• My bad I calculated for x going to inf – Lola May 1 '17 at 15:14
• It yields to 1/3 – Damila Whi Shadow May 1 '17 at 15:19
• Yes that's the answer I found too after using L'H – Lola May 1 '17 at 15:19

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 $(4)$.

$\infty - \infty$ 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 May 1 '17 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 $0$. – Kevin May 1 '17 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 $0$.

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.