Question

In Exercises 41 and \(42,\) find the limit. (Hint: Let \(x=1 / t\) and find the limit as \(t \rightarrow 0^{+}\) ) $$ \lim _{x \rightarrow \infty} x \tan \frac{1}{x} $$

Short answer

The limit of the given function as \(x\) approaches infinity is \(1\).

Step-by-step solution

Change of Variable

Following the hint in the problem, we make a change of variable from \(x\) to \(t\), so given \(x=1 / t\), we substitute this into our limit to obtain: \( \lim _{x \rightarrow \infty} x \tan \frac{1}{x} = \lim_{{t \rightarrow 0^{+}}} \frac{1}{t} \tan t\). The limit is now transformed into \(0 \times \infty\) format.

Rewrite Function to Apply L'Hopital's Rule

To apply L'Hopital's rule, we must write the limit in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) or their equivalent. We can rewrite the limit as \(\lim_{{t \rightarrow 0^{+}}} \frac{\tan t}{t}\). This now falls into a format where we can apply L'Hopital's rule.

Apply L'Hopital's Rule

Using L'Hopital's rule, we can evaluate the limit by taking the derivative of both the numerator and the denominator to obtain: \(\lim_{{t \rightarrow 0^{+}}} \frac{\tan t}{t} = \lim _{{t \rightarrow 0^{+}}} \frac{\frac {d(\tan t)}{dt}}{\frac {dt}{dt}} = \lim _{{t \rightarrow 0^{+}}} \frac{\sec^2 t}{1} = \sec^2(0)\).

Evaluate the Limit

Evaluating the limit, we substitute \(0\) into \(\sec^2 t\) to get \(\sec^2(0) = 1\).

Key concepts

L'Hopital's Rule

L'Hopital's Rule is a fantastic tool in calculus for dealing with indeterminate forms such as \(0/0\) and \(\infty/\infty\). These forms crop up often when evaluating limits, and without a way to address them, we'd be stuck. Fortunately, the rule offers a method to find the limit of a quotient as the numerator and denominator both approach zero or both approach infinity.

The essential idea is to take the derivative of the top (numerator) and bottom (denominator) of the fraction separately and then to evaluate this new limit. L'Hopital's Rule can only be applied when the original limit results in one of the specific indeterminate forms that the rule is designed to handle. Also, it's crucial to ensure the derivatives exist and are continuous in the interval we're investigating.

In the context of the given problem, after transforming the expression, we apply L'Hopital's Rule to \(\lim_{{t \rightarrow 0^{+}}} \frac{\tan t}{t}\) since we faced a \(0 \times \infty\) condition, which we turned into \(\frac{0}{0}\). By differentiating the numerator and the denominator separately, we're able to move forward in solving the limit.

Tangent Function

The tangent function, \(\tan x\), is a periodic function that is one of the basic trigonometric functions. The tangent of an angle in a right triangle is the ratio of the length of the opposite side to the length of the adjacent side. In terms of the unit circle, \(\tan x\) is the ratio of the y-coordinate to the x-coordinate of the corresponding point on the unit circle.

For calculus purposes, the tangent function is especially interesting due to its properties and behavior around certain points. For instance, \(\tan x\) has infinite limits at \(\frac{\pi}{2}+n\pi\), where \(n\) is any integer, since it approaches a vertical asymptote.

When dealing with limits involving the tangent function, like in our exercise, we often exploit its periodicity and symmetry, as well as relationships with other trigonometric functions. In particular, we used the tangent function to convert a limit involving infinity to one that is easier to evaluate at zero after our strategic substitution.

Infinite Limits

Infinite limits refer to situations in calculus where a function's value becomes arbitrarily large (either positively or negatively) as the input approaches a certain point, or as the input itself grows without bound. These types of limits can manifest in many forms—whether it's the input approaching infinity, like \(x \rightarrow \infty\), or approaching a finite value where the function becomes unbounded.

When a problem contains an expression approaching infinity, such as \(x \tan\frac{1}{x}\) as \(x\) goes to infinity, we are looking at an infinite limit. Such problems can be challenging because the behavior of the function may not be straightforward at a glance. This is why we sometimes have to use techniques like substitution or rules like L'Hopital's to transform these expressions into something more manageable.

In the given exercise, the function involves \(x\) approaching infinity, which is an infinite limit. By changing the variable and employing L'Hopital's Rule, we address the complexity introduced by the infinite nature of the limit, bringing the problem into a context where the evaluation of the limit is straightforward.