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

Let's begin by analyzing the given limit problem:

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

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

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

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

Thus, the numerator \(x + \sin(2x)\) approaches \(\infty\) because \(x\) dominates the oscillating \(\sin(2x)\), while the denominator \(x \to +\infty\). 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 \(x\):

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

This simplifies to:

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

### Step 3: Evaluate the Limit

Now we can evaluate the limit term by term:
- \(\lim_{x \to +\infty} 1 = 1\),
- \(\lim_{x \to +\infty} \frac{\sin(2x)}{x} = 0\), since \(\sin(2x)\) is bounded and oscillates between \(-1\) and \(1\), while \(x \to +\infty\) makes the fraction go to 0.

Thus, the overall limit is:

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

### Conclusion:

The limit is:

\[
\boxed{1}
\]

### 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 \(x \to -\infty\)?

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

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

Explore the solution for finding the limit of lim x->+infinity (x + sin(2x)) / x and why L'Hopital's rule doesn't simplify this indeterminate form. Learn how to apply simplification techniques for handling oscillating trigonometric functions, ideal for undergraduate calculus students.

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