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A machine part vibrates so that its displacement is y = x²cos1/x millimetres (mm) at a position x…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A machine part vibrates so that its displacement is $y = x^2\cos\dfrac{1}{x}$ millimetres (mm) at a position $x$ millimetres from a fixed pin, for $x \ne 0$ . The cosine swings faster and faster near the pin. A student wants to show that the displacement dies away at the pin. Find the largest possible value of $|y|$ when $x = 0.1$ mm, using only the fact that the cosine is between $-1$ and $1$ . Then find the limit of $y$ as $x$ tends to $0$ .

Need a reminder? The key ideaShow

Squeeze: if $g \le f \le h$ and $g, h \to L$ , then $f \to L$ . Split $|x - a|$ at $a$ . For the greatest integer, $[x]$ is the largest integer not above $x$ , and $[-3.2] = -4$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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