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

Hence, L Hospital rule for x→c+ is proved. Similarly, we can prove it for x→c.

Indeterminate Forms

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)



\(\lim_{x\rightarrow 0}\) (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


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.

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