L’ Hospital’s rule

L'Hôpital's Rule:

Definition: L'Hôpital's Rule is a powerful tool used to evaluate limits involving indeterminate forms 00 or . It enables the determination of certain limits by taking the derivatives of the numerator and denominator and re-evaluating the limit.

Indeterminate Forms L'Hôpital's Rule Applies To:

  1. 00 form: Occurs when both the numerator and denominator of a function approach zero as the variable approaches a certain value.
  2. form: Arises when both the numerator and denominator of a function grow unbounded as the variable approaches a certain value.

L'Hôpital's Rule Statement: If f(x) and g(x) are differentiable functions on an open interval except possibly at a point c (where c can be a real number or ±), and if limxcf(x)=limxcg(x)=0, then:

limxcf(x)g(x)=limxcf(x)g(x) If the limit on the right-hand side exists or is ±.

Steps for Applying L'Hôpital's Rule:

  1. Verify that the limit is in an indeterminate form of 00 or .
  2. Take the derivatives of the numerator and denominator separately.
  3. Re-evaluate the limit using the derivatives obtained.
  4. Repeat the process if the result still yields an indeterminate form until a determinate value is reached or the limit is proven to be divergent.

Precautions:

  • L'Hôpital's Rule can only be applied when the conditions for its use are met.
  • It doesn’t always guarantee a solution; in some cases, repeated applications might not resolve the limit.

Example: Evaluate the limit limx0sin(x)x.

Solution:

  1. Initial Evaluation: Direct substitution yields limx0sin(x)x=sin(0)0, which is of the form 00, indicating an indeterminate form.

  2. Apply L'Hôpital's Rule:

    • Differentiate the numerator and denominator separately:
      • Derivative of sin(x) is cos(x).
      • Derivative of x is 1.
    • Now, the limit becomes limx0cos(x)1 after applying L'Hôpital's Rule.
  3. Re-evaluate the Limit:

    • Substituting x=0 into cos(x) gives cos(0)=1.
    • Therefore, limx0sin(x)x=limx0cos(x)1=1.

So, using L'Hôpital's Rule, the original indeterminate limit limx0sin(x)x evaluates to 1.