Limit of $f(x)=x^2\left(\sin{\frac 1 x}\right)^2$ for $x\rightarrow\pm\infty$

Question

I have the following function:

$$f(x)=x^2\left(\sin{\frac 1 x}\right)^2$$

For $x\rightarrow\pm\infty$, we have

$$f(x)\sim x^2\cdot\frac{1}{x^2}=1$$

so the limit is $1$. In cases like this I assume that the limit is $1^+$ because every operation I've done to get to that limit involves positive numbers. Therefore I assume that the function approaches $1$ from "above" (not below). However this is not the case, we approach $1$ from below:

The limit is actually $1^-$. Sometimes it is required to draw the graph of a function. How could I have known I had to draw from below $1$? Do I have to look at what the function does in the other points (e.g. for $x\rightarrow$0) Any hints?

Answer 1

Since $\sin(x) \leqslant x$ when is $x$ is enough close to $0$, you have (we take $x > 0$ because of the function parity) :

$$ x^2 \sin(1/x)^2 \leqslant 1$$

And the limit is $1^-$

Answer 2

Since $sin(\frac {1}{x}) \leq \frac {1}{x}$, the graph will always stay below $y=1$.

Answer 3

If you check the first and second derivative of your function for large positive values of x, you figure out that your function is increasing and concave down which gives you a hint about your limit.

Same with large negative values of x.