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

Also Read

→Limits of a Function
→Continuity of a Function
→Differentiability of a Function
→Rules for differentiation
→Higher order derivatives
→L’HOSPITAL’S RULE

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