How do you evaluate #[ (ln x) / (csc x) ]# as x approaches 0+?

Question

Aurora Alvarado · Accepted Answer

#lim_(x->0^+) lnx/cscx = 0#

You can write the function as:

#lnx/cscx = lnx sinx = (x lnx) (sinx/x)#

so that:

#lim_(x->0^+) lnx/cscx = (lim_(x->0^+) (x lnx))* ( lim_(x->0^+)(sinx/x))#

Both limits are well known, but we can remind how they can be calculated:

#lim_(x->0^+)(sinx/x)#

is in the indeterminate form #0/0# and can be calculated using l'Hospital's rule:

#lim_(x->0^+)(sinx/x) = lim_(x->0^+) (d/(dx) sinx)/(d/(dx) x) = lim_(x->0^+) cosx / 1 = 1#

#lim_(x->0^+) (x lnx)#

is in the indeterminate form #0/oo# but can be reduced to a different form by expressing it as a quotient:

#lim_(x->0^+) (x lnx) = lim_(x->0^+) lnx/(1/x)#

This is now in the form #oo/oo# and we apply again l'Hospital's rule:

# lim_(x->0^+) lnx/(1/x) = lim_(x->0^+) (d/(dx) lnx)/(d/(dx) (1/x)) = lim_(x->0^+) (1/x) / (-1/x^2) = lim_(x->0^+) (-x) = 0#

We can then conclude that:

#lim_(x->0^+) lnx/cscx = 0*1 = 0#

Paisley Johnson · Answer

undefined To evaluate [ (ln x) / (csc x) ] as x approaches 0+, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we have [ (1/x) / (-cot x * csc x) ]. Simplifying further, we get [ -cot x / (x * csc x) ]. As x approaches 0+, cot x approaches positive infinity and csc x approaches positive infinity, while x approaches 0. Therefore, the limit of [ -cot x / (x * csc x) ] as x approaches 0+ is negative infinity.