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.  #### 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.

#### 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.

#### 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.

#### 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

#### 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

#### 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.

#### 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