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

Question

Charles Arocha · Accepted Answer

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

1. Take the derivatives of the numerator and denominator separately.
2. Apply L'Hospital's Rule by taking the limit of the ratio of the derivatives as $ x $ approaches 0.
3. 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.

Christopher Agresta · Answer

#lim_(x->0^+)ln(x)/csc(x) = 0# First, assuming we are only consider real numbers, we can only take the limit of #ln(x)# as #x# approaches #0# from the positive direction, and so we will proceed as such. Noting that #lim_(x->0^+)ln(x) = -oo# and #lim_(x->0^+)csc(x) = oo#, we have #lim_(x->0^+)ln(x)/csc(x)# as a #oo/oo# indeterminate form. As such, we can apply L'hopital's rule. #lim_(x->0^+)ln(x)/csc(x) = lim_(x->0^+)(d/dxln(x))/(d/dxcsc(x))# #=lim_(x->0^+)(1/x)/(-csc(x)cot(x))# #=lim_(x->0^+)-sin(x)/x*tan(x)# If two functions converge at a point, then their product also converges at that point, and the limit of the product is the product of the limits of the original functions. Using that fact, together with the known limits #lim_(x->0)sin(x)/x = 1#, and #lim_(x->0)tan(x)=0#, we can factor our expression into the product of two limits: #lim_(x->0^+)ln(x)/csc(x) =lim_(x->0^+)-sin(x)/x*tan(x)# #= -(lim_(x->0^+)sin(x)/x)(lim_(x->0^+)tan(x))# #=-1*0# #=0#