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Chapter 11: Problem 31

Calculate. $$\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)$$

Short Answer

Expert verified
The limit of the given function as x approaches 0 is 0, after algebraic manipulation, applying L'H么pital's Rule twice, and evaluating the resulting simplified expression.

Step by step solution

01

Combine the fractions

First, we need to combine the two fractions under a common denominator, which in this case will be x * sin(x): \[ \left(\frac{1}{\sin{x}} - \frac{1}{x}\right) = \frac{x - \sin{x}}{x\sin{x}} \]
02

Use L'H么pital's Rule

As we try to take the limit, we see it appears as an indeterminate form of type \(\frac{0}{0}\): \[ \lim_{x \to 0} \frac{x - \sin{x}}{x\sin{x}} \] So now we apply L'H么pital's rule (taking derivatives of the numerator and denominator until we have an easier limit to evaluate): \[ \lim_{x \to 0} \frac{1 - \cos{x}}{\sin{x}+x\cos{x}} \]
03

Evaluate the limit

Now we substitute 0 for x in the simplified expression: \[ \frac{1 - \cos{0}}{\sin{0}+0\cos{0}} = \frac{1 - 1}{0 + 0} = \frac{0}{0} \] This still presents an indeterminate form of type \(\frac{0}{0}\), therefore we need to apply L'H么pital's rule again: \[ \lim_{x \to 0} \frac{\sin{x}}{\cos{x}-\sin{x}} \]
04

Evaluate the new limit

Now try substituting 0 for x in the new expression: \[ \frac{\sin{0}}{\cos{0}-\sin{0}} = \frac{0}{1 - 0} = 0 \] So the limit of the original function as x approaches 0 is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L'H么pital's Rule
L'H么pital's Rule is a powerful tool in calculus used to evaluate limits involving indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It's especially helpful when direct substitution in a limit problem doesn't work and results in these forms. L'H么pital's Rule states that if \( \lim_{x \to a} f(x) = 0 \) and \( \lim_{x \to a} g(x) = 0 \), or both tend to \( \infty \), then:\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]provided the limit on the right-hand side exists.
  • L'H么pital's Rule requires differentiable functions in the vicinity of \(x = a\) and assumes the derivatives are continuous.
  • The rule can be applied repeatedly if the resulting expressions still show indeterminate forms.
This concept simplifies complex limits and assists in evaluating expressions that may seem challenging at first glance.
Indeterminate Forms
When dealing with limits, you may encounter expressions that lead to indeterminate forms. These forms occur when the limit becomes unclear and typical arithmetic cannot resolve them directly. Common indeterminate forms include:
  • \(\frac{0}{0}\)
  • \(\frac{\infty}{\infty}\)
  • \(0 \cdot \infty\)
  • \(\infty - \infty\)
  • \(0^0, 1^\infty, \infty^0\)
In the exercise, the indeterminate form \( \frac{0}{0} \) appeared in the limit \( \lim_{x \to 0} \frac{x - \sin{x}}{x\sin{x}} \). To resolve such forms, calculus provides techniques like L'H么pital's Rule or algebraic manipulation to transform the expression into one where direct substitution is feasible.
Trigonometric Limits
Trigonometric limits often occur in calculus due to the involvement of trigonometric functions like \( \sin{x} \), \( \cos{x} \), etc. These functions have specific properties and standard limits that are useful. For trigonometric functions approaching zero, such limits include:
  • \(\lim_{x \to 0} \frac{\sin{x}}{x} = 1\)
  • \(\lim_{x \to 0} \frac{1 - \cos{x}}{x} = 0\)
These trigonometric limits help us rewrite and simplify expressions for limit evaluation. For example, in the present exercise, breaking down \( \sin{x} \) and applying these known limits helped in simplifying the indeterminate form \( \frac{1 - \cos{x}}{\sin{x} + x\cos{x}} \) to eventually find the result. A strong grasp of these limits can make handling complex evaluations easier and more intuitive.

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