Other Indeterminate Forms Help

Evaluate the limit

Solution 1

Notice that x ² → +∞ while e ^{3 x} → 0. So the limit is indeterminate of the form 0 · ∞. We rewrite the limit as

Now both numerator and denominator tend to infinity and we may apply l’Hôpital’s Rule. The result is that the limit equals

Again the numerator and denominator both tend to infinity so we apply l’Hôpital’s Rule to obtain:

It is clear that the limit of this last expression is zero. We conclude that

You Try It : Evaluate the limit .

Example 1

Evaluate the limit

Solution 1

We study the limit of f ( x ) = x ^x by considering ln f ( x ) = x · ln x . We rewrite this as

Both numerator and denominator tend to ±∞, so the quotient is indeterminate of the form −∞/∞. Thus l’Hôpital’s Rule applies. The limit equals

Now the only way that ln f ( x ) can tend to zero is if f ( x ) = x ^x tends to 1. We conclude that

Example 2

Evaluate the limit

Solution 2

Let f ( x ) = (1 + x ² ) ^{ln| x |} and consider ln f ( x ) = ln | x | · ln(1 + x ² ). This expression is indeterminate of the form −∞ · 0.

We rewrite it as

so that both the numerator and denominator tend to 0. So l’Hôpital’s Rule applies and we have

The numerator tends to 0 (see Example 5.3) and the denominator tends to 1. Thus

But the only way that ln f ( x ) can tend to zero is if f ( x ) tends to 1. We conclude that

Example 1

Evaluate the limit

Solution 1

We put the fractions over a common denominator to rewrite our limit as

Both numerator and denominator vanish as x → 0. Thus the quotient has indeterminate form 0/0. By l’Hôpital’s Rule, the limit is therefore equal to

This quotient is still indeterminate; we apply l’Hôpital’s Rule again to obtain

Example 2

Evaluate the limit

Solution 2

The expression is indeterminate of the form ∞ − ∞. We put the two fractions over a common denominator to obtain

Notice that the numerator and denominator both tend to zero as x → 0, so this is indeterminate of the form 0/0. Therefore l’Hôpital’s Rule applies and our limit equals

Again the numerator and denominator tend to zero and we apply l’Hôpital’s Rule; the limit equals

Example 1

Evaluate the limit

Solution 1

The limit as written is of the form ∞ − ∞. We rewrite it as

Notice that both the numerator and denominator tend to zero, so it is now indeterminate of the form 0/0. We may thus apply l’Hôpital’s Rule. The result is that the limit equals

Since this last limit is −2, we conclude that

Example 2

Evaluate

Solution 2

First rewrite the limit as

Notice that both the numerator and denominator tend to zero (here we use the result analogous to Example 5.7 that x ⁴ e ^{3 x} → 0). So our new expression is indeterminate of the form 0/0. l’Hôpital’s Rule applies and our limit equals

Just as in Example 5.7, x ⁴ · e ^{3 x} x ³ · e ^{2 x} , and x ⁴ · e ^{2 x} all tend to zero. We conclude that our limit equals 0.