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

Answer 1

Samantha Adkison

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

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

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

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Answer 2

Elijah Abram

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

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#

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Answer 3

Axel Alvarez

To find the limit of cos(x)/x as x approaches 0 from the positive side, we can use L'Hôpital's Rule. By differentiating both the numerator and denominator, we get -sin(x)/1. Evaluating this expression at x=0, we find that the limit is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find #lim cosx/x# as #x-&gt;0^+#?

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