Math Problem Statement

59—62 Verify that L'Höpital's rule is of no help in finding the limit; then find the limit, if it exists, by some other method.

lim x->+infinity (x+sin2x)/x

Solution

Let's begin by analyzing the given limit problem:

limx+x+sin(2x)x\lim_{x \to +\infty} \frac{x + \sin(2x)}{x}

Step 1: Check if L'Hôpital's Rule is Applicable

L'Hôpital's Rule is applied to limits of the indeterminate forms 00\frac{0}{0} or \frac{\infty}{\infty}. Let's check the form of the limit:

As x+x \to +\infty:

  • x+x \to +\infty,
  • sin(2x)\sin(2x) oscillates between 1-1 and 11, so it stays bounded.

Thus, the numerator x+sin(2x)x + \sin(2x) approaches \infty because xx dominates the oscillating sin(2x)\sin(2x), while the denominator x+x \to +\infty. This gives us the indeterminate form \frac{\infty}{\infty}, which suggests that L'Hôpital's Rule could be applied. However, the rule won't simplify things in this case, as it would require differentiating both the numerator and denominator, but we will show that there’s a simpler way to solve the limit.

Step 2: Simplify the Expression

We can divide both terms in the numerator by xx:

limx+x+sin(2x)x=limx+(xx+sin(2x)x)\lim_{x \to +\infty} \frac{x + \sin(2x)}{x} = \lim_{x \to +\infty} \left( \frac{x}{x} + \frac{\sin(2x)}{x} \right)

This simplifies to:

limx+(1+sin(2x)x)\lim_{x \to +\infty} \left( 1 + \frac{\sin(2x)}{x} \right)

Step 3: Evaluate the Limit

Now we can evaluate the limit term by term:

  • limx+1=1\lim_{x \to +\infty} 1 = 1,
  • limx+sin(2x)x=0\lim_{x \to +\infty} \frac{\sin(2x)}{x} = 0, since sin(2x)\sin(2x) is bounded and oscillates between 1-1 and 11, while x+x \to +\infty makes the fraction go to 0.

Thus, the overall limit is:

limx+(1+sin(2x)x)=1+0=1\lim_{x \to +\infty} \left( 1 + \frac{\sin(2x)}{x} \right) = 1 + 0 = 1

Conclusion:

The limit is:

1\boxed{1}

Would you like further details or have any questions about this solution?

Here are 5 questions to expand the topic:

  1. How does L'Hôpital's rule handle indeterminate forms?
  2. What are the conditions under which L'Hôpital's rule can be applied?
  3. How would the result change if the function had different oscillating terms?
  4. Can you apply the same technique to other trigonometric functions?
  5. What happens if the limit is taken as xx \to -\infty?

Tip: Always check if the expression can be simplified before applying L'Hôpital's Rule. It can often save time!

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Math Problem Analysis

Mathematical Concepts

Limits
Indeterminate forms
Asymptotic behavior
Trigonometric functions

Formulas

lim x->+∞ (x + sin(2x)) / x
L'Hopital's rule for indeterminate forms
sin(2x) oscillation between -1 and 1

Theorems

L'Hopital's Rule
Limit evaluation of trigonometric functions

Suitable Grade Level

Undergraduate Calculus