# How do you find the limit of #y=lnx/cscx # as x approaches 0 using l'hospital's rule?

To find the limit of ( \frac{\ln(x)}{\csc(x)} ) as ( x ) approaches 0 using L'Hospital's Rule, follow these steps:

- Take the derivatives of the numerator and denominator separately.
- Apply L'Hospital's Rule by taking the limit of the ratio of the derivatives as ( x ) approaches 0.
- Repeat the process until you can find the limit or determine if the limit is indeterminate.

Let's begin:

First, find the derivatives of the numerator and denominator:

The derivative of ( \ln(x) ) is ( \frac{1}{x} ). The derivative of ( \csc(x) ) is ( -\csc(x) \cot(x) ).

Now, apply L'Hospital's Rule by taking the limit of the ratio of the derivatives as ( x ) approaches 0:

[ \lim_{x \to 0} \frac{\frac{1}{x}}{-\csc(x) \cot(x)} ]

Now, differentiate again:

The derivative of ( \frac{1}{x} ) is ( -\frac{1}{x^2} ). The derivative of ( -\csc(x) \cot(x) ) can be found using the product rule, which yields ( -\csc(x)(-\csc^2(x) - \cot^2(x)) ).

Now, apply L'Hospital's Rule again:

[ \lim_{x \to 0} \frac{-\frac{1}{x^2}}{-\csc(x)(-\csc^2(x) - \cot^2(x))} ]

Simplify:

[ \lim_{x \to 0} \frac{1}{x^2} \cdot \frac{\csc(x)}{\csc^2(x) + \cot^2(x)} ]

As ( x ) approaches 0, ( \csc(x) ) approaches ( \infty ), and the expression becomes ( \frac{1}{0} \cdot \frac{\infty}{\infty} ), which is an indeterminate form.

Apply L'Hospital's Rule one more time:

[ \lim_{x \to 0} \frac{-\frac{2}{x^3}}{-\csc(x)(-3\csc^2(x)\cot(x)) + 2\csc(x)\csc^2(x))} ]

Now, plug in ( x = 0 ) to get the final result.

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