How do you find the limit of #(x csc x + 1)/(x csc x) # as x approaches 0?

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Answer 1 · 2016-01-29T06:08:49

#2#

Going into this, you should know the following limit identity:

#lim_(xrarr0)sinx/x=1#

This will come in handy.

The question rewritten is:

#lim_(xrarr0)(xcscx+1)/(xcscx)#

Split up the numerator.

#=lim_(xrarr0)1+1/(xcscx)#

#1/(xcscx)# can be simplified since #1/cscx=sinx#.

#=lim_(xrarr0)1+sinx/x#

Limits can be added to one another, as follows:

#=lim_(xrarr0)1+lim_(xrarr0)sinx/x#

Both of these limits are equal to #1#, so the expression equals

#=1+1=2#

We can check a graph of the function. It should approach #2# at #x=0#.

graph{(xcscx+1)/(xcscx) [-14.04, 14.43, -1, 3]}