Informative line

L Hospital Rule

Learn L’Hospital’s rule function and evaluating L’hospital’s Rule by exponential, logarithmic or trigonometric function.

L' Hospital Rule(0/0 form)

Indeterminate form \(\Big(\dfrac {0}{0}\Big)\)

  • A limit of the form \(\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}\) where \(f\) and \(g\)are any two function of \(x\)is said to be of \(\dfrac {0}{0}\)form if \(\lim\limits_{x\to a}\,f(x)=0\) and \(\lim\limits_{x\to a}\,g(x)=0\)
  • When \(f\) and \(g\)are polynomial or algebraic functions the limits can be evaluated by elementary methods but sometimes \(f, \,g\) contains exponential, logarithmic or trigonometric function.
  • For these limits we use L' Hospital Rule which says, that 
    • If \(f\) and \(g\)are differentiable and \(g'(x)\neq0\) on an open interval  \(I\) , that contains \('a'\). Suppose that 

\(\lim\limits_{x\to a}\,f(x)= \lim\limits_{x\to a}\,g(x)=0\) then

\(\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\)

If the limits on the right side exist.

  • So to evaluate \(\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}\)suppose we observe that it is \(\dfrac {0}{0}\) form then we use 

\(\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\)

If \(\dfrac {0}{0}\) form is removed, the limits can be evaluated by direct substitution, otherwise we apply L' Hospital Rule again.

Illustration Questions

Use L' Hospital Rule to evaluate \(\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {1-sin\,\theta}{1+cos2\,\theta}\) . 

A \(\dfrac {1}{4}\)

B \(\dfrac {1}{2}\)

C \(\dfrac {3}{2}\)

D \(\dfrac {3}{4}\)

×

According to L' Hospital Rule

\(\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}\)

if \(f(a)=g(a)=0\)

In this case the limit is

\(\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)\) which is \(\Big(\dfrac {0}{0}\Big)\)form

\(\therefore \;\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)= \lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {\dfrac {d}{d\theta}(1-sin\,\theta)} {\dfrac {d}{d\theta}(1+cos2\,\theta)} \)

\(=\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {-cos\,\theta}{-2sin2\,\theta}= \lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {cos\,\theta} {2\,sin2\,\theta}\)

Put \(\theta\to\dfrac {\pi}{2}\) we get \(\dfrac {0}{0}\) form again

\(\therefore \,\,\lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {cos\,\theta}{2\,sin2\,\theta}= \lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {-sin\,\theta} {4\,cos2\,\theta}= \dfrac {-sin\dfrac {\pi}{2}} {4\,cos\,\pi}=\dfrac {1}{4}\)

\(\therefore \;\;\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)=\dfrac {1}{4}\)

Use L' Hospital Rule to evaluate \(\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {1-sin\,\theta}{1+cos2\,\theta}\) . 

A

\(\dfrac {1}{4}\)

.

B

\(\dfrac {1}{2}\)

C

\(\dfrac {3}{2}\)

D

\(\dfrac {3}{4}\)

Option A is Correct

Indeterminate Form  (0/0 form)

  • A limit of the form \(\lim\limits_{x\to a}\;\dfrac {f(x)}{g(x)}\) where \(f\) and  \(g\)are any two functions of \(x\), is said to be of \(\left (\dfrac {0}{0}\right)\)form if  \(\lim\limits_{x\to a}\;f(x)=0\) and \(\lim\limits_{x\to a}\;g(x)=0\).
  • When \(f\) and  \(g\)are polynomial or algebraic functions, the limit can be evaluated by elementary method, but sometimes \(f\), \(g\) contain exponential, logarithmic and trigonometric functions.
  • For these limits, we use L' Hospital rule.

Illustration Questions

In which one of the following L' Hospital rule can be applied?

A \(\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}\)

B \(\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}\)

C \(\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}\)

D \(\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}\)

×

In option A

\(\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}\)

\(\Rightarrow\;\dfrac {1-sin\,\dfrac {\pi}{2}}{1+cos\,2×\dfrac {\pi}{2}}\)

\(\Rightarrow\;\dfrac {1-1}{1-1}\)

\(\Rightarrow\;\dfrac {0}{0}\) from

Hence, L' Hospital rule can be applied.

Thus option (A) is correct.

In option B

\(\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}\)

\(=\dfrac {1-sin\,0}{1+cos\,0}\)

\(=\dfrac {1}{2}\)

Thus here L' Hospital rule can not be applied as no \(\dfrac{0}{0}\) form is obtained.

Hence, option (B) is incorrect.

In option C

\(\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}\)

\(=\dfrac {1-sin\,0}{1+cos\,0}\)

\(=\dfrac {1}{2}\)

Thus here L' Hospital rule can not be applied as no \(\dfrac{0}{0}\) form is obtained.

Hence, option (C) is incorrect.

In option D

\(\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}\)

\(=\dfrac {1-sin\,\pi}{1+sin\,\pi}\)

\(=\dfrac {1}{1}=1\)

Thus L' Hospital rule can not be applied as no \(\dfrac{0}{0}\) form is obtained.

Hence option (D) is incorrect.

In which one of the following L' Hospital rule can be applied?

A

\(\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}\)

.

B

\(\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}\)

C

\(\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}\)

D

\(\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}\)

Option A is Correct

L' Hospital Rule for Logarithmic Function (infinite by infinite form)

Indeterminate form \(\Big(\dfrac {\infty}{\infty}\Big)\)

  • A limit of the form \(\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}\) where \(f\) and \(g\)are any two function of \(x\)is said to be of \(\Big(\dfrac {\infty}{\infty}\Big)\)form if \(\lim\limits_{x\to a}\,f(x)=\infty\) and \(\lim\limits_{x\to a}\,g(x)=\infty\)
  • When \(f\) and \(g\)are polynomial or algebraic functions the limits can be evaluated by elementary methods but sometimes \(f, \,g\) contains exponential, logarithmic or trigonometric function.
  • For these limits we use L' Hospital Rule which says, that 
    • If \(f\) and \(g\)are differentiable and \(g'(x)\neq \infty\) on an open interval \(I\), that contains \('a'\). Suppose that 

\(\lim\limits_{x\to a}\,f(x)= \lim\limits_{x\to a}\,g(x)=\infty\) then

\(\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\)

if the limits on the right side exist.

  • So to evaluate \(\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}\)suppose we observe that it is \(\Big(\dfrac {\infty}{\infty}\Big)\) form then we use 

\(\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\to\)If \(\Big(\dfrac {\infty}{\infty}\Big)\) form is removed the limits can be evaluated by direct substitution, otherwise we apply L' Hospital Rule again.

Illustration Questions

Use L' Hospital Rule to evaluate \(\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)\) .

A \(\dfrac {1}{2}\)

B \(0\)

C \(\ell n\,2\)

D \(1\)

×

According to L' Hospital Rule

\(\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}\)

if \(f(a)=g(a)=\infty\)

In this case the limit is

\(\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)\) which is \(\Big(\dfrac {\infty}{\infty}\Big)\)form

\(\therefore \; \lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big) = \lim\limits_{x \to \infty}\;\dfrac {\dfrac {d}{dx}(\ell n\,x)} {\dfrac {d}{dx}(\sqrt x)} \)

\(=\lim\limits_{x\to \infty}\; \dfrac {\dfrac {1}{x}} {\dfrac {1}{2}x^{-1/2}}\)

\(=\lim\limits_{x\to\infty}\;\dfrac {2\sqrt x}{x}= \lim\limits_{x\to\infty}\;\dfrac {2}{\sqrt x}=0\)

\(\therefore \; \lim\limits_{x\to\infty}\;\dfrac {\ell n\,x}{\sqrt x}=0\)

Use L' Hospital Rule to evaluate \(\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)\) .

A

\(\dfrac {1}{2}\)

.

B

\(0\)

C

\(\ell n\,2\)

D

\(1\)

Option B is Correct

Indeterminate Product Form

Indeterminate Form\((0\times \infty)\)

  • \(\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)\) is said to be of the form \(0×\infty\). If \(\lim\limits_{x\to a}\,f(x)=0\) and \(\lim\limits_{x\to a}\,g(x)=\infty\) (or \(-\infty\)).
  • If the above (limit cannot be evaluated by elementary methods we find expression \(\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)\) as \(=\lim\limits_{x \to a}\dfrac {f(x)}{1/g\,(x)}\) or \(\lim\limits_{x \to a}\dfrac {g(x)}{1/f\,(x)}\) and then use L' Hospital rule.

Illustration Questions

Evaluate \(\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)\) .

A \(\dfrac {2}{\pi}\)

B \(\dfrac {-2}{\pi}\)

C \(\dfrac {\pi}{2}\)

D \(\dfrac {-\pi}{2}\)

×

\(\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)=\lim\limits_{x \to a}\dfrac {f(x)}{1/g\,(x)}\) or \(\lim\limits_{x \to a}\dfrac {g(x)}{1/f\,(x)}\)

and then apply L' Hospital rule (for \(0×\infty\) form)

In this case the limit is

\(\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)= \lim\limits_{x \to 1}\;\dfrac {(\ell n\,x)}{\left (cot\,\dfrac {\pi\,x}{2}\right)}\)

which is of \(\left (\dfrac {0}{0}\right)\)form.

\(\therefore \lim\limits_{x \to 1}\;\dfrac {(\ell n\,x)}{\left (cot\,\dfrac {\pi\,x}{2}\right)} \)

\(= \lim\limits_{x \to 1}\;\dfrac {\dfrac {d}{dx}(\ell n\,x)}{\dfrac {d}{dx}\left (cot\,\dfrac {\pi\,x}{2}\right)}\)

\(= \lim\limits_{x \to 1}\;\dfrac {\dfrac {1}{x}}{-cosec^2\,\dfrac {\pi\,x}{2}×\dfrac {\pi}{2}}\)

\(=\dfrac {-2}{\pi}\)

\(\therefore \lim\limits_{x\to 1}\,(\ell n\,x)\Big(tan\,\dfrac{\pi\,x}{2}\Big)=\dfrac {-2}{\pi}\)

Evaluate \(\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)\) .

A

\(\dfrac {2}{\pi}\)

.

B

\(\dfrac {-2}{\pi}\)

C

\(\dfrac {\pi}{2}\)

D

\(\dfrac {-\pi}{2}\)

Option B is Correct

Illustration Questions

Use L' Hospital Rule to evaluate \(\lim\limits_{x \to 0}\;\dfrac {x\,2^x}{2^x-1}\) .

A \(\dfrac {1}{2\, \ell n\,2}\)

B \(2\,\ell n\,2\)

C \(\ell n\,2\)

D \(\dfrac {1}{\ell n\,2}\)

×

According to L' Hospital Rule

\(\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}\)

if \(f(a)=g(a)=0\)

In this case the limit is

\(\lim\limits_{x \to 0}\;\left (\dfrac {x\,2^x}{2^x-1}\right)\) which is \(\Big(\dfrac {0}{0}\Big)\)form

\(\therefore \; \lim\limits_{x \to 0}\;\left (\dfrac {x\,2^x}{2^x-1}\right)= \lim\limits_{x \to 0}\; \dfrac {\dfrac {d}{dx}(x\,2^x)}{\dfrac {d}{dx}(2^x-1)}\)

\(= \lim\limits_{x \to 0}\; \dfrac {x×2^x\,\ell n\,2+2^x}{2^x\,\ell n\,2}\) (Now put \(x = 0\) )

\(=\dfrac {0+1}{\ell n\,2}=\dfrac {1}{\ell n\,2}\)

Use L' Hospital Rule to evaluate \(\lim\limits_{x \to 0}\;\dfrac {x\,2^x}{2^x-1}\) .

A

\(\dfrac {1}{2\, \ell n\,2}\)

.

B

\(2\,\ell n\,2\)

C

\(\ell n\,2\)

D

\(\dfrac {1}{\ell n\,2}\)

Option D is Correct

Illustration Questions

Use L' Hospital Rule to evaluate \(\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x} \right)\) .

A \(-\pi\)

B \(-\dfrac {1}{\pi^2}\)

C \(\dfrac {1}{\pi^2}\)

D \(\pi\)

×

According to L' Hospital Rule

\(\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}\)

if \(f(a)=g(a)=0\)

In this case the limit is

\(\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right)\) which is \(\Big(\dfrac {0}{0}\Big)\)form

\(\therefore \; \lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right) = \lim\limits_{x \to 1}\; \dfrac {\dfrac {d}{dx}(1-x+\ell n\,x)}{\dfrac {d}{dx}(1+cos\,\pi\,x)}\)

\(=\lim\limits_{x \to 1}\;\left (\dfrac {-1+1/x}{-\pi\,\sin\,\pi\,x}\right)\)

Put \(x=1\), we get  \(\Big(\dfrac {0}{0}\Big)\)form again

\(\therefore \;\lim\limits_{x \to 1}\;\left (\dfrac {-1+1/x}{-\pi\,\sin\,\pi\,x}\right)\)

\(=\lim\limits_{x \to 1}\;\dfrac {\dfrac {d}{dx}(-1+1/x)} {\dfrac {d}{dx}(-\pi\,\sin\,\pi\,x)}\)

\(=\lim\limits_{x \to 1}\; \dfrac {\dfrac {-1}{x^2}}{-\pi^2\,cos\,\pi\,x}=\dfrac {-1}{\pi^2}\)

\(\therefore, \;\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right)=\dfrac {-1}{\pi^2}\)

Use L' Hospital Rule to evaluate \(\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x} \right)\) .

A

\(-\pi\)

.

B

\(-\dfrac {1}{\pi^2}\)

C

\(\dfrac {1}{\pi^2}\)

D

\(\pi\)

Option B is Correct

Indeterminate Power Forms

The Limits of the form \(\lim\limits_{x\to a}\;\Big(f(x)\Big)^{g(x)}\) can be of any one of the following indeterminate form

  1. \(1^{\infty}\to\) when \(\lim\limits_{x\to a}\; f(x)=1\) and \(\lim\limits_{x\to a}\; g(x)=\infty\)
  2. \(0^0\to\) when \(\lim\limits_{x\to a}\; f(x)= \lim\limits_{x\to a}\; g(x)=0\)
  3. \(\infty^0\to\)form when \(\lim\limits_{x\to a}\; f(x)=\infty\) & \(\lim\limits_{x\to a}\; g(x)=0\)

Each of the these limits can be evaluated by assuming the limits to be y and the taking log on both sides and applying L' Hospital rule. 

 

Illustration Questions

Evaluate \(\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}\) using L' Hospital Rule.

A \(e^{(-2/\pi)}\)

B \(e^{(2/\pi)}\)

C \(e^{(\pi/2)}\)

D \(e^{(-\pi/2)}\)

×

For evaluating \(\lim\limits_{x \to a}\,\big(f(x)\big)^{g(x)}\) we first take log and then use L' Hospital Rule.

Let 

\(y=\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}\) which is of \(1^{\infty}\) form.

Take log on both sides \(\Rightarrow\)\(log\,y=\lim\limits_{x\to 1}\; {\Big(tan\dfrac {\pi\, x}{2}\Big)}\;\big(\ell n\,(2-x)\big)\)

\(\Rightarrow\) \(log\,y=\lim\limits_{x\to 1}\; \dfrac {\ell n\,(2-x)}{{\Big(cot\dfrac {\pi\, x}{2}\Big)}\;}\)

Apply L' Hospital Rule

\(log\,y=\lim\limits_{x\to 1}\; \dfrac {\dfrac {d}{dx}\ell n\,(2-x)} {\dfrac {d}{dx}{\Big(cot\dfrac {\pi\, x}{2}\Big)}\;}\)

\(=\lim\limits_{x\to 1}\; \dfrac {\dfrac {1}{2-x}×(-1)} {-cosec^2\,\dfrac {\pi\,x}{2}×\dfrac {\pi}{2}\;}\)

\(=\dfrac {-1}{-1×\dfrac {\pi}{2}}=\dfrac{2} {\pi}\)

\(\Rightarrow\;y=e^{(2/\pi)}\)

Evaluate \(\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}\) using L' Hospital Rule.

A

\(e^{(-2/\pi)}\)

.

B

\(e^{(2/\pi)}\)

C

\(e^{(\pi/2)}\)

D

\(e^{(-\pi/2)}\)

Option B is Correct

Practice Now