L Hospital Rule

In Calculus, the most important rule is L’ Hospital’s Rule (L’Hôpital’s rule). This rule uses the derivatives to evaluate the limits which involve the indeterminate forms. In this article, we are going to discuss the formula and proof for the L Hospital’s rule along with examples.

What is L Hospital Rule?

L Hospital rule is a general method of evaluating indeterminate forms such as 0/0 or ∞/∞. To evaluate the limits of indeterminate forms for the derivatives in calculus, L Hospital’s rule is used. L Hospital rule can be applied more than once. You can apply this rule still it holds any indefinite form every time after its applications. If the problem is out of the indeterminate forms, you can’t be able to apply L Hosptial Rule.

L Hospital Rule Formula

L’ hospital’s rule states that

If \(\lim_{x\rightarrow c}f(x)= \lim_{x\rightarrow c}g(x)=0 \ or\pm \infty\) and \(\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}\), then

\(\lim_{x\rightarrow c}\frac{f(x)}{g(x)}= \lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}\)

Note: The limit of the quotient of function is equivalent to the limit of the quotient of their derivatives, given that the provided conditions are satisfied.

L Hospital Rule proof

By the use of Extended Mean Value Theorem or Cauchy’s Mean Value Theorem, the L Hospital rule can be proved.

If f and g are two continuous functions on the interval [a, b] and differentiable on the interval (a, b), the

f’(c)/g’(c) = [f(b)-f(a)]/[g(b)-g(a)] such that c belong to (a, b).

Assume that the two functions f and g are defined on the interval (c, b) in such a way that f(x)→0 and g(x)→0, as x→c+.

But we have f’(c) / g’(c) tends to finite limits. The functions f and g are differentiable, and f’(x) and g’(x) exists on the set [ c, c+k], and also f’ and g’ are continuous on the interval [c, c+k] provided with the conditions f(c)= g(c) = 0 and g’(c) ≠ 0 on the interval [c, c+k].

By Cauchy Mean Value Theorem states that there exists ck∈ (c, c+k), such that

f’(ck)/g’(ck) = [f(c+k)-f(c)]/[g(c+k)-g(c)] = f(c+k)/g(c+k)

Now, k→0+,

\(\lim_{k\rightarrow 0^{+}}\frac{f'(c_{k})}{g'(c_{k})}=\lim_{x\rightarrow c^{+}}\frac{f'(x)}{g'(x)}\)

While, \(\lim_{k\rightarrow 0^{+}}\frac{f(c+k)}{g(c+k)}=\lim_{x\rightarrow 0^{+}}\frac{f(x)}{g (x)}\)

So, we have \(\lim_{x\rightarrow c^{+}}\frac{f(x)}{g(x)}= \lim_{x\rightarrow c^{+}}\frac{f'(x)}{g'(x)}\)

L Hospital Rule Uses

Using L Hospital rule, we can solve the problem in 0/0, ∞/∞, ∞ – ∞, 0 x ∞, 1∞, ∞0, or 00 forms. These forms are known as indeterminate forms. To remove the indeterminate forms in the problem, we can use L Hospital rule.

L Hospital Rule Examples

Some examples of L Hospital rule are given below:

Example 1:

Evaluate \(\lim_{x\rightarrow 0}\) (2 sin x – sin 2x) / (x – sin x)

Solution:

Given:

\(\lim_{x\rightarrow 0}\) (2 sin x – sin 2x) / (x – sin x)

Differentiate the above form, we get

= \(\lim_{x\rightarrow 0}\) (2 cos x – 2 cos 2x) / (1 – cos x)

= \(\lim_{x\rightarrow 0}\) (-2 sin x + 4 sin 2x) / (sin x)

= \(\lim_{x\rightarrow 0}\) (-2 cos x + 8 cos 2x) / (cosx)

Now substitute the limit,

= -2+8/1 = 6/1 = 6

Therefore, \(\lim_{x\rightarrow 0}\) (2 sin x – sin 2x) / (x – sin x) = 6.

Example 2:

Evaluate \(\lim_{x\rightarrow 0}\) sin 3x/ sin 4x

Solution:

Given: \(\lim_{x\rightarrow 0}\) sin 3x/ sin 4x

= \(\lim_{x\rightarrow 0}\) 3cos 3x /4 cos 4x

Now substitute the limit,

= 3 cos 0/ 4 cos 0

= ¾

Therefore, \(\lim_{x\rightarrow 0}\) sin 3x/ sin 4x = ¾.

Register with BYJU’S – The Learning App for Maths-related concepts with examples, and also watch engaging videos.