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

2 Answers
Shwetank Mauria
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

Justin
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

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