Indeterminate Forms and L’Hospital Rule (Solved Problems)

Indeterminate Form and L’Hospital Rule: L’Hôpital’s rule in Calculus is a powerful tool for solving indeterminate limits of the forms like 0/0 or ∞/∞ by taking the derivatives of the top and bottom simultaneously and then re-evaluating the limit of the resulting quotient. This rule is repeatedly applied till the limit exists, and this value is equal to the original limit.

Indeterminate Forms of Limits

An indeterminate form of limits is a kind of limit where one cannot get a specific value after putting the limit. For example, lim_x→0 $\dfrac{\sin x}{x}$ is of the form $\dfrac{0}{0}$, and it is an indeterminate form; because 0/0 has no definite value.

Types of Indeterminate Forms

The seven common indeterminate forms in Calculus (Limits) are given as follows:

1. 0/0
2. ∞/∞
3. 0 × ∞
4. ∞ – ∞
5. 1^∞
6. 0⁰
7. ∞⁰

These indeterminate forms of limits can be solved using L’Hospital Rule.

L’Hospital Rule: Solved Problems

Type 0/0

Q1: Find the limit $\lim \limits_{x \to 0} \dfrac{1-\cos x}{x^2}$.

Answer:

Step 1: The direct substitution gives us $\dfrac{1-\cos 0}{0^2}=\dfrac{0}{0}$, so the given limit is an indeterminate form 0/0. Now, in the next step, we will apply L’Hospital rule.

Step 2: lim_x→0 $\dfrac{1-\cos x}{x^2}$

= lim_x→0 $\dfrac{(1-\cos x)’}{(x^2)’}$ where $’$ denotes the first order derivative.

= lim_x→0 $\dfrac{\sin x}{2x}$

= $\dfrac{1}{2} \times$ lim_x→0$ \dfrac{\sin x}{x}$

Step 3: Use the fact lim_x→0 sinx/x =1. So the above limit is equal to

= $\dfrac{1}{2} \times 1$

= $\dfrac{1}{2}$.

So the value of the limit lim_x→0 (1-cosx)/x² is equal to 1/2, and it is obtained by the L’Hospital Rule.

Q2: Find the limit $\lim \limits_{x \to 1} \dfrac{\ln x}{x-1}$.

Answer:

Step 1: The direct substitution gives $\dfrac{\ln 1}{1-1}=\dfrac{0}{0}$, so it is an indeterminate form 0/0. Applying L’Hospital rule to it, we have:

Step 2: lim_x→1 $\dfrac{\ln x}{x-1}$

= lim_x→1 $\dfrac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x-1)}$

= lim_x→1 $\dfrac{\frac{1}{x}}{1}$

= lim_x→1 $\dfrac{1}{x}$

= 1/1 = 1.

= $\dfrac{1}{2}$.

So the value of lim_x→1 $\dfrac{\ln x}{x-1}$ is equal to 1.

Q3: Find $\lim \limits_{x \to 0} \dfrac{x-\sin x}{x^3}$.

Answer:

As sin0=0, so the given limit is a 0/0 form. Applying L’Hospital rule to it, we have:

lim_x→0 $\dfrac{x-\sin x}{x^3}$

= lim_x→0 $\dfrac{(x-\sin x)’}{(x^3)’}$

= lim_x→0 $\dfrac{1-\cos x}{3x^2}$

= $\dfrac{1}{3} \times$ lim_x→0 $\dfrac{1-\cos x}{x^2}$

= $\dfrac{1}{3} \times \dfrac{1}{2}$, by Q1 above.

= 1/6.

So the value of lim_x→0 $\dfrac{x-\sin x}{x^3}$ is equal to 1/6.

Type ∞/∞

Q4: Find $\lim \limits_{x \to \infty} \dfrac{x}{e^x}$.

Answer:

The given limit is ∞/∞ form. Therefore,

lim_x→∞ $\dfrac{x}{e^x}$

= lim_x→∞ $\dfrac{\frac{d}{dx}(x)}{\frac{d}{dx}(e^x)}$

= lim_x→∞ $\dfrac{1}{e^x}$

= 0.

So the value of x/e^x as x approaches infinity is equal to 0.

Type 0⁰

Q5: Find limx→0+ $x^x$.

Answer:

The given limit is of type 0⁰. Therefore,

Let y = lim_x→0+ $x^x$

Taking logarithms on both sides, we have that

ln y = lim_x→0+ $\ln(x^x)$

⇒ ln y = lim_x→0+ $x \ln x$

⇒ ln y = lim_x→0+ $\frac{\ln x}{1/x}$

Using L’Hospital rule, ln y = lim_x→0+ $\frac{1/x}{-1/x^2}$

⇒ ln y = lim_x→0+ $(-x)$ = 0

⇒ y = e⁰ = 1.

So the value of the limit of x^x as x approaches 0 is equal to 1.

Type ∞-∞

Q6: Find limx→0 $\left( \dfrac{1}{x}-\dfrac{1}{\sin x}\right)$.

Answer:

The given limit is of type ∞-∞. Therefore,

lim_x→0 $\left( \dfrac{1}{x}-\dfrac{1}{\sin x}\right)$

= lim_x→0 $\dfrac{\sin x -x}{x\sin x}$ (0/0 form)

= lim_x→0 $\dfrac{(\sin x -x)’}{(x\sin x)’}$

= lim_x→0 $\dfrac{\cos x -1}{\sin x+x \cos x}$ (0/0 form)

= lim_x→0 $\dfrac{-\sin x}{\cos x+ \cos x-x \sin x}$

= $\dfrac{-\sin 0}{\cos 0+ \cos 0-0 \cdot \sin 0}$

= $\dfrac{-0}{1+ 1-0}$ = 0.

Q7: Find $a$ and $b$ for which the limit exists.

Answer:

Practice Problems

Using L’Hospital rule or otherwise, find the following limits:

1. lim_x→0 x cotx
2. lim_x→0 $\left(\dfrac{1}{x^2} -\dfrac{1}{\tan x} \right)$

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