When working with limits of the form , we can use the exponential function to turn the limit into the form . Both and approach , so we can apply L'Hopital's rule: However, this is still ∞/∞ so we apply L'Hopital's rule again: However, this brings us back to where we started, so we need to use another method to evaluate the limit. Therefore we can apply L'Hopital's rule to to get: Plugging these into the previous formula: We can simplify the right-hand side to get: Now we plug this back into the previous equation to get: Remembering our original rewrite that , we know that: Notice that we can cancel out a term from the left and right side of the above equation to get: Now, if we multiply both sides by , we get L'Hopital's rule for the ∞/∞ case. But first, since and we can rewrite the above as: This is now of the form 0/0 so we can use L'Hopital's rule: Both and approach infinity, so we can call the limit L for now and take the exponential of both sides. This is the same limit we evaluated in Example 4, so. We write exp(x) for exso to reduce the amount exponents. For case 1, the limit is called an intermediate form of type 0/0. But this is still a limit of the form ∞/∞, and we would have to apply L'Hopital's rule 1000 times to be able to evaluate the limit. If g wins, the result is 1. In the case of a tie, the limit will be a finite number. For limits of the form 0/0, if the numerator wins, then the limit will be 0. This is why we should always plug the value that x is approaching into the limit to make sure that there is not an easier way to evaluate the limit before using L'Hopital's rule. If we plug in x = -4, we get 0/0, so when we apply L'Hopital's rule we get: Note: Instead of using L'Hopital's rule, we could have multiplied both top and bottom by , which is the conjugate of the numerator. lim x→∞ (1 + 1 x)x= exp(ln( lim x→∞ (1 + 1 x)x)) = exp( lim x→∞ ln((1 + 1 x)x)) = exp( lim x→∞ xln(1 + 1 x)) = exp( lim x→∞ ln(1 + 1 x) 1/x) We can now apply L’Hopital’s since the limit is of the form 0 0. We may have expected this because as x approaches ∞, , which can be approximated as . Since the original limit was 2, L'Hopital's rule does not apply. is unimaginably large, ex grows to infinity so the limit is still +∞. Recall the statement of L'Hopital's rule: If f and g are differentiable and g'(x) ≠ 0 on an open interval containing a (except possibly at a) and one of the following holds: given that the limit on the right side exists or is . Since the exponential is a continuous function like ln, we can exchange the order of the exponential and limit operations: We could use L'Hopital's rule at this point since both and x approach +∞ so the limit is of type ∞/∞, but the math would be messy. The second is an \({\infty }/{\infty }\;\) indeterminate form, but we can’t just factor an \({x^2}\)out of the numerator. For these indeterminate forms that involve exponents such as 1∞, 00, ∞0, we need to use the natural log function to turn the limit into the form 00 or ∞∞ so that we can use L'Hopital's rule (see the trick in Implicit Differentiation for an example of how we use the ln function). Alternatively, we could have noted that and rewritten. We choose between 0/0 and ∞/∞ based on which is easier to compute. If there is a tie, the result is a finite number. If there is a tie, the result is a finite number. If there is a tie, then the limit will be a finite number as in the 0/0 case. If g wins, the result is ∞. Whenever and , the limit is called an indeterminate form of type 1 ∞. Similarly, in case 2, is called an indeterminate of the form ∞/∞. As x grows large, the limit is of the form ∞0, so for now, we call the limit and take the ln of both sides to get: Since ln is a continuous function, we can exchange the order of ln and lim symbols to get: Since is just the reciprocal of the first example, , so L = 1. and so this limit is of type 1∞ and we need to take the ln of this limit: Now, this is in the form 0/0, so we apply L'Hopital's rule: Since this is the ln of the original limit, the original limit must be e0 = 1. To see why for the 0/0 case, recall the definition of a derivative at the point x = a: If we assume that f' and g' are continuous at a, then their limits near x = a equal their values at x = a: Dividing these two equations and remembering that from the properties of limits: Notice that we can cancel out the x - a terms on the right-most side to get: Since we assumed that is type 0/0, we can also say that. L'Hopital's rule is a theorem that can be used to evaluate difficult limits. In a way, this is the reverse technique of using the ln function to evaluate indeterminate forms of type 1∞, ∞0, and 00. Although L'Hopital's rule can sometimes be used with indeterminate limits of other forms, it is most typically useful for limits of the forms mentioned: 0/0 or ∞/∞. 00 case: If f wins, the result is 0. If we apply L'Hopital's rule we get: which tends to infinity, so the limit is . Namely, for limits of type ∞/∞, if the numerator wins, the limit will be ∞. By L'Hopital's rule: which is of the form ∞/∞, but solving this with L'Hopital's rule would be more complicated. This equals , so we can guess that the limit should be . Infinity, negative or positive, over zero will always result in divergence. Also, completing the square tells us that . "Winning," as it is used here and throughout the rest of the article, refers to which part of the function is dominant, i.e., which one is reaching its limit faster. After doing so, we would obtain: whose limit at -4 can be evaluated by plugging in x = -4. This is where the subject of this section comes into play. Finally, while limits resulting in zero, infinity, or negative infinity are often indeterminate forms, this is not always true. As x approaches ∞, both x and lnx approach infinity, so this is an indeterminate limit of type ∞/∞. Also, L'Hopital's rule does not always work because in some cases, repeatedly applying L'Hopital's rule will still result in indeterminate forms regardless of how many times the rule is applied.
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