L'Hospital's Rule

L'Hospital's rule (read: lopital's rule) can sometimes be used to calculate limits in the form of proportions.

Definition

If we are calculating the limit of a function at a point, then we can use derivatives to help us in certain circumstances. So let's have a share function, i.e. we have the functions f(x) and g(x) and we are looking for the limit of

$$ \lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}. $$

If the equations f(x₀) = g(x₀) = 0 hold and there is a limit

$$ \lim_{x\rightarrow x_0}\frac{f'(x)}{g'(x)}, $$

then the relation holds:

$$ \lim_{x\rightarrow x_0}\frac{f(x)}{g(x)} = \lim_{x\rightarrow x_0}\frac{f'(x)}{g'(x)}. $$

If the limit of $\lim_{x\rightarrow x_0}\frac{f'(x)}{g'(x)}$ does not exist, then it does not follow that the limit of $\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}$ does not exist either. If there is no limit with derivations, we have to compute the original limit without derivations in a different way.

Example

1. Calculate the limit

   $$ \lim_{x\rightarrow 3}\frac{x^2-9}{x^2-x-6} $$

   Can we solve this example with L'Hospital's rule? First, we need to check that the functional value of the two functions (the one in the denominator of the fraction and the one in the numerator) at the point x₀ = 3 is zero. Let us denote the functions f(x) = x² − 9 and g(x) = x² − x − 6. Then:

   $$ \begin{eqnarray} f(3) &=& 3^2 - 9 = 0\\ g(3) &=& 3^2 - 3 - 6 = 0 \end{eqnarray} $$

   The first condition is satisfied. Note that we cannot calculate the limit simply by putting the number three after x, because the whole function is discontinuous at the point x₀ = 3, or does not have a functional value there at all (it does not have a functional value because we would have to divide by zero to get it). Now we can try to use L'Hospital's rule and derive the two functions. Note that we are not deriving by the formula for the quotient, we are deriving separately the function in the numerator and the function in the denominator. So we get:

   $$ \begin{eqnarray} f'(x) &=& 2x\\ g'(x) &=& 2x - 1 \end{eqnarray} $$

   So instead of the previous limit, we can solve this limit:

   $$ \lim_{x\rightarrow3}\frac{2x}{2x-1} $$

   This function is continuous at the point x₀ = 3, so we can just substitute the value of three after x and calculate the functional value of the whole function:

   $$ \lim_{x\rightarrow3}\frac{2x}{2x-1} = \frac{2\cdot3}{2\cdot3-1}=\frac65 $$

   We can see that the limit of this function exists, therefore the limit of $\lim_{x\rightarrow 3}\frac{x^2-9}{x^2-x-6}$ also exists and these limits are equal. Thus:

   $$ \lim_{x\rightarrow 3}\frac{x^2-9}{x^2-x-6} = \lim_{x\rightarrow3}\frac{2x}{2x-1} = \frac65. $$

2. Calculate the limit

   $$ \lim_{x\rightarrow0}\frac{x}{\sin x} $$

   What is the situation at x₀ = 0? Again, denote f(x) = x and g(x) = sin x. It is true that f(0) = 0 and g(0) = 0. We get the proportion $\frac00$, the function is discontinuous at that point. We can try to use L'Hospital's rule. We combine the two functions:

   $$ \begin{eqnarray} f'(x) &=& 1\\ g'(x) &=& \cos x \end{eqnarray} $$

   We have the derivation of the functions, let's try to calculate the limit

   $$ \lim_{x\rightarrow0}\frac{1}{\cos x} $$

   What is the situation at the point x₀ = 0? Since cos(0) = 1, we get the fraction $\frac11=1$. The function is continuous at the point x₀ = 0, so the following holds

   $$ \lim_{x\rightarrow0}\frac{1}{\cos x}=\frac11=1. $$

   Since this limit exists, there is a limit $\lim_{x\rightarrow0}\frac{x}{\sin x}$ and these limits are equal. We get

   $$ \lim_{x\rightarrow0}\frac{x}{\sin x} = 1 $$