# Why can't we apply the L'Hopital rule once the function is no longer an indeterminate form?

As I understand it, L'Hopital Rule comes from Cauchy's Generalised Mean Value Theorem, which states that if 2 functions, $f(x)$ and $g(x)$ are continuous and differentiable over the interval $[a,b]$ and $(a,b)$ respectively, then $∃$ $c∈[a,b]$ such that $$(f(b)-f(a))/(g(b)-g(a))=(f'(c))/(g'(c))$$

Now if $f(x)$ and $g(x)$ are arbitrary functions, why shouldn't L'Hopital Rule hold once the function is no longer indeterminate? Is there something I didn't see in the proof?

• Do you mean that $b\to a$ (to ensure that $c\to a$)? – A.Γ. Aug 5 '18 at 7:52
• Consider the limit of $\frac{3x^2}{2x}$ where x is tends to 1.Now putting $x=1$ we have the value $3/2$.But if we apply L'Hospital it will give us wrong answer! – Sufaid Saleel Aug 5 '18 at 7:54
• @A.Γ. I'm not sure what you mean. From my understanding of Cauchy's Mean Value Theorem, a and b are boundaries you define for a function, and c is a point on the function where the relationship stated applies. – Yip Jung Hon Aug 5 '18 at 7:59
• @SufaidSaleel Yes, that's true. But I don't understand why you have to stop applying L'Hopital Rule once the function is determinate. – Yip Jung Hon Aug 5 '18 at 8:00
• @YipJungHon L'Hopital rule apples to limits. It is not clear what limit that you mean in the question. – A.Γ. Aug 5 '18 at 8:05

When $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$, then we can apply Cauchy's theorem:$$\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-f(a)}=\frac{f'(c)}{g'(c)},$$for some $c$ and therefore$$\lim_{x\to a}\frac{f'(x)}{g'(x)}=L\implies\lim_{x\to a}\frac{f(x)}{g(x)}=L.$$But if $\lim_{x\to a}f(x)\neq0$ or $\lim_{x\to a}g(x)\neq0$, this argument doesn't apply anymore.

Of course, this doesn't actually prove that we can't apply L'Hopital's rule then. What proves that is the existence of counter examples.

I usually motivate L'hospital with Taylor series (or linear approximation if it's early in the course.) See that (using $a=0$ for convenience)

$$\lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{f(0) +f'(0)x+f''(0)x^2/2+\cdots}{g(0)+g'(0)x+g''(0)x^2/2+\cdots}.$$

If both $f(0)=0$ and $g(0)=0$, this becomes

$$\lim_{x\to 0} \frac{f'(0)x+f''(0)x^2/2+\cdots}{g'(0)x+g''(0)x^2/2+\cdots}.$$

But if they are not both $0$, then I can't do this step. Then I can cancel one $x$ in the fraction to get

$$\lim_{x\to 0} \frac{f'(0)+f''(0)x/2+\cdots}{g'(0)+g''(0)x/2+\cdots},$$

and we can repeat this reasoning until we hit a derivative that's not $0$ at $0$.