L’Hospital’s Rule

L'Hospital's rule

Let f(x) and g(x) be two functions differentiable in the neighbourhood of the point 'a', except at the point 'a' itself. If $\underset{x\to a}{\lim },$ f(x)=$\underset{x\to a}{\lim }$g(x) = 0. Or, $\underset{x\to a}{\lim }$f(x) = $\underset{x\to a}{\lim }$g(x) = $\infty$. Then$\underset{x\to a}{\lim }$$\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{fapos\left(x\right)}{gapos\left(x\right)}$provided that the limit on the right exist as a finite number or is $\pm \infty$Illustration1: If $\underset{x\to a}{\lim }\frac{a^x-x^a}{x^x-a^a}$ = – 1 then find the value of a.Solution Since the given limit is in the form of 0/0, we will use L’ Hospital’s Rule $\underset{x\to a}{\lim }\frac{a^x-x^a}{x^x-a^a}$ = $\underset{x\to a}{\lim }\frac{a^x\log a-a{x}^{a-1}}{x^x+x^x{\log }_{e}x}=-1$$\frac{a^a\log a-a{a}^{a-1}}{a^a+a^a\log a}=\frac{\log a-1}{\log a+1}=-1$

It is satisfied only when a = 1.

Note:

$\bullet$L'Hospitals rule is not always useful. Consider the example,$\underset{x\to \infty }{\lim }\frac{x+\sin x}{x-\sin x}$$\left(form\frac{\infty }{\infty }\right)$.Here, if we apply $Lapos$ Hospital’s rule, then $\underset{x\to \infty }{\lim }\frac{x+\sin x}{x-\sin x}=\underset{x\to \infty }{\lim }\frac{1+\cos x}{1-\cos x}$Now, both the N^r and the D^r are undefined because $\underset{x\to \propto }{\lim }$ cos x doesn’t existWe can find this limit as follows:$\underset{x\to \infty }{\lim }\frac{x+\sin x}{x-\sin x}=\underset{x\to \infty }{\lim }\frac{1+\left(\frac{\sin x}{x}\right)}{1-\left(\frac{\sin x}{x}\right)}=\frac{1+0}{1-0}=1$(since $\underset{x\to \infty }{\lim }$$\frac{\sin x}{x}$ = 0)