Question

Question #11ebf

Answer 1

1

Explanation:

So we know that on the real numbers, `sin` and `cos` are bounded. This means that

`-1<=cos(x)<=1` and `-1<=sin(x)<=1` `AA` `x in RR`

So we can manipulate the limit to take advantage of this. We start with

`lim_(xrarroo) (x^2 + sin(x))/(x^2+cos(x))`

Now divide top and bottom by `x^2`, the limit becomes:

`lim_(xrarroo) (1 + (sin(x))/(x^2))/(1+(cos(x))/(x^2))`

As x tends to infinity `sin(x)` and `cos(x)` remain bounded however `x^2 rarr oo`, hence

`lim_(xrarroo) sin(x)/x^2 = 0` and `lim_(xrarroo) cos(x)/x^2 = 0`

So we have that

`lim_(xrarroo) (1 + (sin(x))/(x^2))/(1+(cos(x))/(x^2)) = 1/1 = 1`