Question

In Exercises 2.4.2-2.4.40, find the indicated limits. $$ \lim _{x \rightarrow-\infty} e^{x^{2}} \sin \left(e^{x}\right) $$

Short answer

Answer: The limit as \(x\) goes to negative infinity of the expression \(e^{x^2}\sin (e^x)\) is 0.

Step-by-step solution

Analyze the exponential function

As x goes to negative infinity, the exponent in the exponential function, \(x^2\), will become larger and larger (since the square of a large negative number becomes a large positive number). As a result, the value of the exponential function \(e^{x^2}\) goes to infinity as \(x\) approaches negative infinity.

Analyze the sine function

As x goes to negative infinity, the input to the sine function, \(e^x\), will go to zero (since \(e^x\) approaches 0 as \(x\) goes to negative infinity). The sine function oscillates between -1 and 1, so as its input goes to 0, the value of \(\sin(e^x)\) approaches \(\sin(0) = 0\).

Multiply the two functions

Now, we need to consider the limit as \(x\) goes to negative infinity of the product of these two functions. As \(x\) approaches negative infinity, the exponential function \(e^{x^2}\) goes to infinity, and the sine function \(\sin(e^x)\) goes to 0. We need to determine how this product behaves as \(x\) approaches negative infinity.

Apply the limit

The limit of the product of two functions is the product of their limits, provided that both limits exist. We have: $$ \lim _{x \rightarrow-\infty} e^{x^{2}} \sin \left(e^{x}\right) = \left(\lim_{x \rightarrow-\infty} e^{x^2}\right) \cdot \left(\lim_{x \rightarrow-\infty} \sin(e^x)\right) $$ Since we know \(\lim_{x \rightarrow-\infty} e^{x^2} = \infty\) and \(\lim_{x \rightarrow-\infty} \sin(e^x) = 0\), we need to determine the value of the limit as \(x\) approaches negative infinity of this product. As \(x\) approaches negative infinity, we have an infinite factor (\(e^{x^2}\)) multiplied by a factor approaching 0 (\(\sin(e^x)\)). When we multiply infinity by 0, this is an indeterminate form. To solve for the limit, we can apply L'Hôpital's rule or rewrite the expression into a different form in order to gain more insight.

Rewrite the expression

Let \(y = e^{x}\). Then, as \(x\) goes to negative infinity, \(y\) goes to 0. Therefore, the limit becomes: $$ \lim _{y \rightarrow 0} \frac{\sin y}{\frac{1}{y^2}} $$

Apply L'Hôpital's Rule

Now, the expression is in the indeterminate form of \(\frac{0}{0}\). We can use L'Hôpital's rule by taking the derivatives of both the numerator and the denominator with respect to \(y\). We have: $$ \lim_{y \rightarrow 0} \frac{\cos y}{-2/y^3} = \lim_{y \rightarrow 0} \frac{-y^3\cos y}{2} $$ As \(y\) goes to 0, we know that \(\cos y\) goes to 1. Thus, the limit becomes: $$ \lim_{y \rightarrow 0} \frac{-y^3\cos y}{2} = \frac{-0^3 \cdot 1}{2} = 0 $$

Conclusion

The limit as \(x\) goes to negative infinity of the given expression is 0. Thus, we have: $$\lim _{x \rightarrow-\infty} e^{x^{2}} \sin \left(e^{x}\right) = 0$$

Key concepts

Indeterminate Forms

When dealing with limits in real analysis, we often encounter scenarios where the limit expression doesn't lead to a straight-forward answer. These situations are known as indeterminate forms. A classic example is when the limit seems to result in \(0 \times \infty\), like in the exercise you're studying. In these cases, because infinity is not a real number and zero times any real number is zero, the limit is inconclusive as stated and requires further manipulation or special techniques to resolve.

Other common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0^0\), \(\infty^0\), and \(1^\infty\). Recognizing these forms is crucial because it signals to the problem-solver that direct substitution will not work, and they must employ other methods, like factorization, simplification, or advanced calculus techniques such as L'Hôpital's rule, to find the limit.

L'Hôpital's Rule

When you're stuck with an indeterminate form, L'Hôpital's rule is your mathematical superhero. This rule allows you to evaluate the limit of a fraction when both the numerator and the denominator approach 0 or \(\infty\). The trick is to take the derivative of the numerator and the denominator separately and then find the limit of the new fraction. This process can be repeated as needed if the resulting limit is still indeterminate.

It's important to note, however, that L'Hôpital's rule can only be used under specific conditions: both the numerator and denominator must be differentiable near the point of interest, and after taking the derivatives, you should end up with a determinable limit. If these criteria are not met, then this rule isn't applicable, and other strategies must be used.

Exponential Functions

Exponential functions like \(e^{x}\) play a significant role in mathematics, and especially in calculus. These functions are characterized by a constant base raised to a variable exponent. When dealing with limits that involve exponential functions, there are key behaviors to remember. As \(x \to \infty\), the function \(e^{x}\) grows rapidly towards infinity. Conversely, as \(x \to -\infty\), the function \(e^{x}\) approaches 0.

Exponential growth is incredibly fast, much faster than polynomial growth. That's why, when faced with limits where an exponential function is raised to a variable square, as in \(e^{x^2}\) for large negative values of \(x\), you see results heading towards infinity rather than zero, despite the negative exponent.

Sine Function Behavior

The \(\sin\) function is one of the basic trigonometric functions and has a well-understood behavior when approaching limits. Its values oscillate between -1 and 1. The limit of the \(\sin\) function as \(x\) approaches any real number is simply the \(\sin\) of that number. This predictability simplifies the limit evaluation process, particularly when \(x\) approaches 0 because \(\sin(0) = 0\).

However, while \(\sin(x)\) has a direct limit as \(x\) approaches any real number, when \(\sin(x)\) is multiplied by another function that grows without bound, it creates more complexity. Since the behavior of the \(\sin\) function is consistent and bounded, it often plays a more passive role in shaping the outcome of such limits. The unbounded growth of its partner function tends to dominate the behavior of the product.