How do you find #lim cosx/x# as #x->0^+#?

2 Answers
Jun 17, 2017

#Lt_(x->0^+)cosx/x=oo#

Explanation:

In #Lt_(x->0^+)cosx/x#, as #x->0#, #cosx->cos0=1# and #x->0#

Hence #Lt_(x->0^+)cosx/x=1/0=oo#

Jun 21, 2017

#lim_(x->0^+)cosx/x=+oo#

Explanation:

Apart from using the method shown by the other contributor, which is just plugging in 0 and finding that it approaches #oo#, there is another, more sophisticated method of showing it, which is to use the Taylor approximation of #cosx# as #x->0#, or otherwise known as the Maclaurin expansion of #cosx#.

The Maclaurin expansion of #cosx# is #1-x^2/2+x^4/(4!)-x^6/(6!)+...#

Plugging in the Maclaurin expansion into the limit gives:

#lim_(x->0^+)cosx/x=lim_(x->0^+)(1-x^2/2+x^4/(4!)-x^6/(6!)+...)/x#

Simplifying gives:

#lim_(x->0^+)1/x-x/2+x^3/(4!)-x^5/(6!)+...#

When #x# tends to 0, all the terms from the 2nd onwards become 0.
Therefore, the only term left is the first term, which is #lim_(x->0^+)1/x#. This leaves us with #+oo#, hence
#lim_(x->0^+)cosx/x=+oo#