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# L'Hopital's Rule For Solving Indeterminate Limits Last updated date: 29th Nov 2023
Total views: 21.3k
Views today: 0.21k     ## L'Hopital's Rule For Computing Indeterminate Forms and Limits

L'Hopital's Rule enables us to simplify the evaluation/assessment of limits that involve indeterminate forms. An indeterminate form is described as a limit that does not provide sufficient information to recognize the original limit. The frequently appearing indeterminate forms that take place in calculus and other aspects of Mathematics include:

00, ∞∞, 1∞, ∞−∞, 00, ∞0, 0×∞

L'Hopital's Rule can be stated in following the following manner:

Assuming that f and g are both differentiable functions that address 0 (or ±∞) as x→a. Then the function f(x), and g(x) approaches the same limit as f′(x) and, g′(x) respectively.

The derivatives on both the denominator as well as the numerator are taken individually.

### Topics Covered in L’Hôpital’s Rule

• Identify when to apply L’Hôpital’s Rule

• Recognize indeterminate forms generated by products, powers, subtractions, and quotients, and how to apply L’Hôpital’s rule in each case.

• Explain the relative growth rates of functions.

### Recognizing Limits Using L’Hôpital’s Rule

The problems stated like these involve the use of L'Hopital's Rule. L’Hôpital’s Rule is applied to bypass the common indeterminate forms ”∞"∞"∞"∞ and "0"0"0"0 and when calculating limits. You might be surprised to know that there are innumerable forms of L"Hopital's Rule, whose verifications need cultivated techniques in calculus, and can be found in many calculus books. Given are the two forms of L’Hopital's Rule.

### Theorem 1: L’Hopital's Rule for 0 over 0

Let’s assume that lim x→mf [x] = 0 lim x→mf[x]= 0, limx→mg(x)=0limx→mg[x]=0, and that functions ff and gg are distinguishable on an open interval II consisting of mm. Also, suppose that g′(x) ≠ 0g′[x] ≠0 in II if x≠mx≠m.

Then,

limx→mf [x] g[x] =limx→mf ′[x]g′[x] limx→mf [x] g[x] = limx→mf′ [x]g′[x]

As long as the limit is finite, +∞+∞, or −∞−∞. Same outcomes bear for x→∞ x→∞ and x→−∞ x→−∞.

### Theorem 2: L’Hopital's Rule for ∞ over ∞ {Infinity Over Infinity}

Let’s suppose that functions ff and gg are distinguishable for all xx greater than some fixed number. If limx→mf[x] = ∞limx→mf[x] = ∞

and limx→mg[x] = ∞limx→mg[x] = ∞,

then,

limx→mf [x] g [x] = limx→mf′[x]g′[x]limx→mf[x]g(x) = limx→mf′ [x] g ′[x]

as long as the limit is finite- −∞−∞ OR +∞+∞. Same outcomes bear for x→∞x→∞ and x→−∞x→−∞.

In both these forms of L'Hopital's Rule, you must remember that you are required to distinguish (individually) the numerator and denominator of the ratio if either of the indeterminate forms (zero & infinite) "0"0"0"0 or "∞"∞"∞"∞occurs in the calculation of a limit.

Note: Do not get confused between the L'Hopital's Rule and the Quotient Rule for derivatives as the two are not similar. These two are different aspects.

1. ### limx→2 [x2+x−6x2−4]

Solution:

At x=2, we would generally obtain:

22+2− 6× 22− 4 = 0

Now, we will apply to L'Hôpital Rule

Check for derivative rules and differentiate both top and bottom

lim x→2 [x2+x−6x2−4]

= lim x→2 [2x+1−02x−0]

Now we will just exchange x = 2 to obtain our answer:

limx→2 [2x+1−02x−0] = 54

## FAQs on L'Hopital's Rule For Solving Indeterminate Limits

Q1. What is the condition in L'Hospital of limits?

Answer: As of Differentiable, for a limit addressing C, the original functions should be differentiable at either side of c, but not compulsorily at c.

Similarly make a note that g’(x) will not be equal ≠ to 0 at either side of c.

This limit should exist:

limx→cf’ [x] g’[x]

You might be wondering why? Well an excellent example is functions that never settle to a value.

### limx→∞ x + cos[x]x

Which is a clear case of ∞∞

Now, let's differentiate top and bottom:

limx→∞1−sin[x] 1

And since it only quivers up and down, it is unable to ever address any value.

So that NEW limit does not exist!

Thus, L'Hôpital's Rule is not applicable in this condition.

However, we alternatively perform an action like below:-

limx→∞x+cos(x)x = limx→∞[1 + cos[x][x]

As x proceeds further to infinity then cos[x] x is likely to be between −1∞ and +1∞, and both are disposed to zero.

And we are left with just the "1", thus we get:

limx→∞x+cos [x]x = limx→∞ [1 + cos{x}x] = 1

Q2. How was L'Hopital's discovered?

Answer: Although popular as L'Hospital's Rule, the theorem was first discovered by a Swiss Mathematician— Johann Bernoulli in 1694 and possibly introduced to L'Hospital. The publication of the textbook of L'Hospital on differential calculus was released in 1696, the original premier, included the theorem and thus L'Hopital's name got linked to the rule.