Evaluate lim x->0+ ((ln x)/(x-1) - csc x), without using any approximation (is l hopital rule applicable)

To evaluate the limit 

\[
\lim_{x \to 0^+} \left( \frac{\ln x}{x - 1} - \csc x \right),
\]

we first rewrite it for clarity:

\[
\lim_{x \to 0^+} \left( \frac{\ln x}{x - 1} - \frac{1}{\sin x} \right).
\]

Both terms tend toward \(-\infty\) as \(x \to 0^+\) (since \(\ln x \to -\infty\) and \(\sin x \to 0\)). Thus, the limit presents an indeterminate form of \(-\infty - (-\infty)\), which simplifies to an indeterminate form. 

We can combine the fractions:

\[
\frac{\ln x}{x - 1} - \csc x = \frac{\ln x}{x - 1} - \frac{1}{\sin x} = \frac{\ln x \sin x - (x - 1)}{(x - 1) \sin x}.
\]

Now, we will evaluate the limit:

\[
\lim_{x \to 0^+} \frac{\ln x \sin x - (x - 1)}{(x - 1) \sin x}.
\]

### Step 1: Evaluate the numerator

As \(x \to 0^+\):

- \(\ln x \to -\infty\)
- \(\sin x \approx x\) (as \(x \to 0\)), thus \(\ln x \sin x \to -\infty\)

The numerator \(\ln x \sin x - (x - 1)\) also tends to \(-\infty\) since \((x - 1) \to -1\).

### Step 2: Evaluate the denominator

As \(x \to 0^+\):

- \(x - 1 \to -1\)
- \(\sin x \to 0\)

Thus, \((x - 1) \sin x \to 0\).

### Step 3: Apply L'Hôpital's Rule

Since we have the form \(\frac{-\infty}{0}\), we apply L'Hôpital's Rule, differentiating the numerator and denominator:

1. **Differentiate the numerator**:
   - \(f(x) = \ln x \sin x - (x - 1)\)
   - Using the product rule, \(f'(x) = \frac{1}{x} \sin x + \ln x \cos x - 1\).

2. **Differentiate the denominator**:
   - \(g(x) = (x - 1) \sin x\)
   - Using the product rule, \(g'(x) = \sin x + (x - 1) \cos x\).

### Step 4: Compute the limit again

Now we compute:

\[
\lim_{x \to 0^+} \frac{\frac{1}{x} \sin x + \ln x \cos x - 1}{\sin x + (x - 1) \cos x}.
\]

Substituting \(x \to 0^+\):

- \(\frac{1}{x} \sin x \to 0\),
- \(\ln x \cos x \to -\infty\),
- The numerator thus approaches \(-\infty\).

In the denominator:

- \(\sin x \to 0\),
- \((x - 1) \cos x \to -1\),
- So the denominator approaches \(-1\).

Thus, we again have:

\[
\frac{-\infty}{-1} = \infty.
\]

### Final Result

The limit evaluates to 

\[
\lim_{x \to 0^+} \left( \frac{\ln x}{x - 1} - \csc x \right) = \infty.
\]

If you have any questions or need further details, feel free to ask! 

### Further Questions:
1. What is the significance of L'Hôpital's Rule in this context?
2. Can you explain how the behavior of \(\ln x\) near zero influences the limit?
3. How does the Taylor series expansion of \(\sin x\) affect the evaluation of limits?
4. What alternative methods could be used to evaluate this limit?
5. What are some other indeterminate forms similar to this one? 

### Tip
When evaluating limits that involve logarithmic or trigonometric functions, always check for common indeterminate forms and consider algebraic manipulation or L'Hôpital's Rule for clarity.

Explore the evaluation of the limit lim x->0+ ((ln x)/(x-1) - csc x) using L'Hôpital's Rule. This solution addresses the indeterminate form and provides a step-by-step analysis, suitable for advanced high school students.

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