$\lim_{x \to \infty} x\sin(\frac{1}{x}) = 1$ (epsilon-delta like condition) So, it is clear in many ways that this limit holds.  However, I am interested in proving this with an "epsilon-delta" like condition.  The start of my proof:
Let $f(x) = x\,\sin\left(\frac{1}{x}\right)$.  We would like to show that for any $\epsilon > 0$, there exists an $\alpha > 0$ such that $x >\alpha$ implies $|f(x)-1| < \epsilon$.  But observe that 
$$|f(x)-1| = \left|x\,\sin\left(\frac{1}{x}\right) - 1\right| \leq \left|x\,\sin\left(\frac{1}{x}\right)\right| + 1 \leq |x| + 1 = x + 1\,\,\,.$$
This is where I am not seeing how to proceed.  How can we choose $\alpha$ such that $x > \alpha$, but yet $x + 1 < \epsilon$ ?  Any guidance would be much appreciated (without the Taylor theorem and relevant series expansions).

$\displaystyle \left|x\,\text{sin}\left(\frac{1}{x}\right) - 1\right| \leq \left|x\,\text{sin}\left(\frac{1}{x}\right)\right| + 1$ is true but not helpful, as the left hand side is about $0$ and the right about $2$.
For $x\gt 0$ you have $\dfrac1x \gt \sin\left(\dfrac1x\right) \gt \dfrac1x-\dfrac1{6x^3}$ and so: $$1 \gt x\sin\left(\dfrac1x\right) \gt 1-\dfrac1{6x^2}.$$  
Now consider what happens as $x \to \infty$ using the squeeze theorem.  
If $x \gt \sqrt{\dfrac1{6\epsilon}}$ then $\left|x\,\text{sin}\left(\frac{1}{x}\right) - 1\right| \lt \epsilon$.
