Question

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

Answer 1

`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`

Answer 2

`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`