91Ó°ÊÓ

When evaluating limits, we sometimes encounter indeterminate forms like \( \frac{0}{0} \). In such cases, **L'Hôpital's Rule** can often come to the rescue. It states that if the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can compute the limit as \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided this new limit exists.

Here's how it works:

• Differentiate the numerator \( f(x) \) to get \( f'(x) \).
• Differentiate the denominator \( g(x) \) to get \( g'(x) \).
• Take the limit of the new fraction \( \frac{f'(x)}{g'(x)} \).

In the exercise, when evaluating \( \lim_{x \to -\pi} \frac{1+\cos x}{x+\pi} \), the limit initially gives \( \frac{0}{0} \), an indeterminate form. By applying L'Hôpital's Rule, we differentiate the top to get \(-\sin x\) and the bottom to get \(1\). The new limit \( \lim_{x \to -\pi} \frac{-\sin x}{1} \) equals zero, which resolves the indeterminate form easily.

In calculus, the concept of **exponential decay** is crucial when evaluating limits involving exponential functions. Exponential decay means a quantity decreases at a rate proportional to its current value, which is common in physical and natural processes.

The term \( e^{-x} \) describes such decay. As \( x \rightarrow \infty \), \( e^{-x} \rightarrow 0 \). It decays faster than any polynomial increases, even as \(x\) becomes indefinitely large.

For example, in the limit \( \lim_{x \to \infty} x^2 e^{-x} \), the polynomial term \( x^2 \) grows, but not as quickly as the exponential term decays to zero. The exponential decay dominates, driving the entire expression toward zero. So the limit is zero. Remember:

• Exponential terms usually outweigh polynomial terms when \( x \rightarrow \infty \).
• Fast decay to zero is a characteristic of exponential functions like \( e^{-x} \).

This makes exponential terms powerful in determining limits.

The **Squeeze Theorem** is a handy tool in calculus for finding limits. It works by "squeezing" a function between two other functions whose limits are known and equal as \( x \) approaches a point. If the squeezed function lies between these two functions, then it must also approach the same limit.

Imagine three functions, \( f(x) \), \( g(x) \), and \( h(x) \), where \( f(x) \leq g(x) \leq h(x) \). If both \( f(x) \) and \( h(x) \) tend to the same limit \( L \) as \( x \to c \), then \( g(x) \) must tend to \( L \) as well.

In the problem \( \lim_{x \to 0} \frac{\sin x}{x} \), we can use the Squeeze Theorem:

• We know \(-1 \leq \sin x \leq 1\).
• Multiplying by \( \frac{1}{x} \), we get \(-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}\).
• As \( x \to 0 \), both \(-\frac{1}{x}\) and \(\frac{1}{x}\) approach \(0\), so \( \lim_{x \to 0} \frac{\sin x}{x} = 1\) in line with the theorem.

This principle can help to find limits when a direct evaluation is tricky.

A **vertical asymptote** is a line that a graph approaches but never actually reaches. This occurs when a function’s value increases or decreases without bound as it approaches a particular \( x \) value.

Take, for example, the limit \( \lim_{x \to 1^{-}} \frac{1}{x-1} \). Here, as \( x \) approaches 1 from the left side, the expression \( x-1 \) becomes a very small negative number. This makes the expression \( \frac{1}{x-1} \) drop towards \(-\infty\).

To identify when there’s a vertical asymptote:

• Check if the denominator approaches zero, causing undefined behavior.
• Analyze the direction of approach—whether from left or right.
• Observe the function’s behavior as it tends to infinity or negative infinity.

In such cases, we say there is a vertical asymptote at \( x = 1 \), where the graph keeps moving closer to but never actually touches or crosses this line.