The hospital rule in calculus works very well in the evaluation of limits whose value is an indeterminate value after the direct application of the limit. To apply this rule, we simply replace the given fraction of functions with the fraction of their derivatives, and then apply the limit. We can always apply this rule as often as necessary. Let`s look at how we apply the Hoptial rule to the indeterminate form of zero on zero. Direct application of x → ∞ results in ∞/∞. We are therefore applying the rule again. Then we get: Solution: The numerator and denominator both go to 0$, so we have the right to use the Hospital rule: $$lim_{xrightarrow 0},{ x over e^{2x }-1}= lim_{xrightarrow 0},{ 1 over 2e^{2x }}$$ In the new expression, numerator and denominator are both nonzero if $x=0$, so we just plug $0, To get $$lim_{xrightarrow 0}, { x over e^{2x }-1}= lim_{xrightarrow 0},{ 1 over 2e^{2x }}={1over 2e^0}={1over 2}$$ Solution: $0^+$ means that we are approaching $0$ on the positive side, otherwise we do not have a real value logarithm. This problem illustrates the possibility and necessity of reorganizing a border into a report of things to legitimately apply the Hospital rule. Here we reorganize $$lim_{xrightarrow 0^+},x,ln x=lim_{xrightarrow 0}{ln,xover 1/x}$$ In the new expressions, the top goes to $-infty$ and the bottom goes to $+infty$, while $x$ goes to $0$ (from the right). Therefore, we are entitled to apply the L`Hospital rule and get $$lim_{xrightarrow 0^+},x,ln x=lim_{xrightarrow 0}{ln,xover 1/x}=lim_{xrightarrow 0}{1/xover -1/x^2}$$.

Now it is very necessary to rearrange the expression in the last limit: We have $$lim_{xrightarrow 0}{1/xover -1/x^2}= lim_{xrightarrow 0},- x$$ The new expression is very easy to evaluate: the limit is $0$. The Rule of The Hospital has different names such as the Rule of The Hospital, the Rule of the Hospital, the Rule of Bernoulli, etc. and is used to evaluate the limits of indeterminate forms. It was first introduced in 1694 by the Swiss mathematician Johann Bernoulli and is therefore known as Bernoulli`s rule. It was later developed by a French mathematician Guillaume de l`Hôpital and thus became popular under the name L`Hospitalal rule. If we apply the L`Hospital rule several times, simplify the rational expression each time before applying the limit each time. Otherwise, we will lead to an incorrect answer. Here is an example. Let us look at the rule of the Hospital with its statement, proofs and examples. Let`s also look at some common misconceptions that can arise when applying this rule.

There are cases where we tend to apply the Hospital rule, but this leads to an erroneous result. We must therefore be careful in applying this rule. Let`s see where we can go wrong in the application of the rule. Sometimes the limit leads to an indeterminate form even after a single application of the rule of Hospitality. In this case, we can apply the same rule again and again as necessary. Example 1: Evaluate the limit lim x → ∞ (2×2 + 5x + 3) / (3×2 – 7x + 2) (i) without use and (ii) using the rule of Hospitality. Yes, we can apply the rule of L`Hopital several times. However, before each rule is applied, make sure that applying the limit results in an indeterminate value such as 0/0. In addition, I would like to point out that there will be times when The Hospital`s rule will have to be applied more than once to successfully calculate the limit. To apply the rule of L`Hopital to evaluate a limit lim x → a f (x) / g (x): We can apply the rule of L`Hospital, also commonly known as the rule of L`Hôpital, if the direct replacement of a limit gives an indeterminate form.

It is important to note that the hospital rule treats f(x) and g(x) as independent functions and is not the application of the quotient rule. Solution: Both the numerator and denominator have a limit of $0$, so we have the right to apply the L`Hospital rule: $$lim_{xrightarrow 0},{sin,xover x}= lim_{xrightarrow 0},{cos,xover 1}.$$ In the new expression, neither numerator nor denominator is $0$ to $x=$0, and we can simply plug in to see that the limit is $1. By directly applying the limit x → 0, we obtain (1 – cos 0) / 0 = (1 – 1) / 0 = 0 / 0, which is an indeterminate form. We therefore apply the rule of The Hospital. We know that the derivatives of 1 – cos 2x and x2 are 2x and 2x respectively. Then the above limit is: To evaluate a limit as x → a, we first replace x = a in the specified expression. If this gives an indeterminate value such as 0/0, ∞/∞, ∞/0, etc., we apply the rule of Hospitality. Solution: The numerator and denominator both go to $infty$ as $xrightarrow +infty$, so we have the right to apply the L`Hospital rule to convert this to $$lim_{xrightarrow +infty}{2xover e^x}$$.

But the numerator and denominator always go to $infty$, so reapply L`Hospital`s rule: the limit is $$lim_{xrightarrow +infty}{2over e^x}=0$$, since now the numerator is fixed while the denominator goes to $+infty$. The formula of the L-Hopitals rule is lim x → a f(x) / g(x) = lim x → a f` (x) / g`(x), where the limit on the left side gives an indeterminate form by applying the limit x = a. Here, f`(x) and g`(x) are the derivatives of the respective functions. It states that if we divide one function by another, the limit is the same after taking the derivative of each function (with some special conditions that will be shown later). ☛ So apply the limit first and make sure that an indeterminate shape is created before applying the rule. The indefinite form is something that cannot be defined mathematically.