Question

Evaluate the limit as x tends to infinity `(cos x+sin x)/x^2`?

Answer 1

`lim_(x->0) (cosx+sinx)/x^2 = +oo`

Explanation:

#lim_(x->0) (cosx+sinx)/x^2 = lim_(x->0) cosx/x^2 + lim_(x->0) sinx/x^2#

Considering `lim_(x->0) cosx/x^2 = 1/0 -> +oo`

Considering `lim_(x->0) sinx/x^2 = lim_(x->0) (sinx/x * 1/x)`

`= lim_(x->0) sinx/x * lim_(x->0) 1/x`

`= 1 * 1/0 -> +oo`

Hence, `lim_(x->0) (cosx+sinx)/x^2 = +oo`

We can infer this result from the graph of `(cosx+sinx)/x^2` as `x->0` from above and below.

graph{ (cosx+sinx)/x^2 [-18.02, 17.99, -9, 9.02]}