# 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 may be at the point ‘ a ’ itself . If

$lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x) = 0$

or, $lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x) = \infty$

Then , $lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}$ ; provided that the limit on the right exist as a finite number or is ± ∞

Illustration: Evaluate :

(i) $lim_{x \rightarrow 1} (1-x^2)log_{x^2}2$

(ii) $lim_{x \rightarrow \pi/2} \frac{sinx -(sinx)^{sinx}}{1-sinx + ln sinx}$

(iii) $lim_{x \rightarrow \pi/4} (2-tanx)^{\frac{1}{ln tanx}}$

(iv) $\lim_{x \rightarrow y} \frac{x^y – y^x}{x^x – y^y}$

(v) $lim_{x \rightarrow 1} (tan^{-1}(x)\frac{4}{\pi})^{\frac{1}{x^2-1}}$

(vi) $lim_{x \rightarrow 0} \frac{log_{sec(x/2)}cosx}{log_{secx}cos(x/2)}$

Solution: (i) $lim_{x \rightarrow 1} \frac{1-x^2}{log_2 x^2}$

$lim_{x \rightarrow 1} \frac{1-x^2}{2 log_2 x}$

$lim_{x \rightarrow 1} \frac{1-x^2}{2 \frac{ln x}{ln 2}}$

$lim_{x \rightarrow 1} (\frac{1-x^2}{2 ln x}) ln 2$

Applying L’hospital’s Rule ,

$lim_{x \rightarrow 1} (\frac{-2 x^2}{2 })ln 2 = – ln 2$

(ii) Let sinx = t , t → 1

$lim_{x \rightarrow 1^-} \frac{t -t^t}{1 – t + ln t}$

Applying L’ Hospital Rule ;

$lim_{x \rightarrow 1^-} \frac{1 -t^t (1+ln t)}{0 – 1 + \frac{1}{t}}$

$lim_{x \rightarrow 1^-} \frac{0 -\{t^t (\frac{1}{t}) + t^t(1 + ln t)^2\} }{0 – \frac{1}{t^2}}$

= 2