Collection of Maths Problems

Limits requiring a bit of thought

Task number: 2986

• Variant 1

  \(\displaystyle \lim_{x\to 0} \frac{1}{x^2}. \)

• Result

  The limit exist and is \(\infty\).

• Variant 2

  \(\displaystyle \lim_{x\to 0} \frac{1}{x^3}. \)

• Result

  The limit does not exist, because \[ \lim_{x\to 0^+} \frac{1}{x^3} = \infty \] and \[ \lim_{x\to 0^-} \frac{1}{x^3} = -\infty. \]

• Variant 3

  \(\displaystyle \lim_{x \to \infty} \hbox{tan} \left( \frac{\pi}2 + \frac{1}{x^2 + 4}\right). \)

• Resolution

  Let \(g(x) = \frac{\pi}2 + \frac{1}{x^2 + 4}\) a \(f(y) = \hbox{tan}(y)\). We would like to compute \(\lim_{x \to \infty} f(g(x))\). We can easily compute that \[ \lim_{x \to \infty} g(x) = \frac{\pi}2. \] We'd like to use the theorem about the limit of a composite function, but the problem is that \[ \lim_{y \to \frac{\pi}2} \hbox{tan}(y) \] does not exist.

  Does this mean that the limit we are seeking does not exist? Not necessarily – we do not know this yet! We've only determined that we can't directly use the theorem about the limit of a composite function.

  To resolve this example, we need to realize that \(g(x) > \frac{\pi}2\) for all \(x\) in some neighborhood of \(\infty\). (In this particular example this inequality holds everywhere.) From this it follows (after checking the conditions for the limit of a composite function) that \[ \lim_{x \to \infty} f(g(x)) = \lim_{y \to \frac{\pi}2^+} f(y) = \lim_{y \to \frac{\pi}2^+} \hbox{tan}(y) = -\infty. \]

  It only remains to verify one of the conditions for the limit of a composite function. Specifically, that \(g(x)\) never has the value \(\frac{\pi}2\) in any (punctured) neighborhood of \(\infty\). But actually \(g(x)\) never has the value \(\frac{\pi}2\).

• Variant 4

  \(\displaystyle \lim_{x \to 0} \sin \left(\frac 1x\right). \)

• Hint

  Hint: Consider the sequences \(x_k = \frac{1}{2k\pi}\) and \(y_k = \frac{1}{\pi/2 + 2k\pi}\). We have \(\lim_{k \to \infty} x_k = \lim_{k \to \infty} y_k = 0\). But \[ \lim_{x \to 0} \sin \left(\frac 1{x_k}\right) \] and \[ \lim_{x \to 0} \sin \left(\frac 1{y_k}\right) \] turn out to be different. (Try drawing a picture.)

• Result

  The limit does not exist.

• Variant 5

  \(\displaystyle \lim_{x \to 0} x \sin \left(\frac 1x\right). \)

• Hint

  Give the function an upper bound of \(|x|\) and a lower bound of \(-|x|\).

• Result

  \(0\)

• Variant 6

  \(\displaystyle \lim_{x \to 0} \frac{\sin\left(x \sin \left(\frac 1x\right)\right)}{x \sin \left(\frac 1x\right)}. \)

• Resolution

  We'd like to use the theorem about the limit of a composite function. Let \(g(x) = x \sin \left(\frac 1x\right)\) and \(f(y) = \frac{\sin y}{y}\). So we'd like to compute \(\lim_{x \rightarrow 0} f(g(x))\).

  From the preceding variations of this exercise we know that \(\lim_{x \to 0} g(x) = 0\) and also we have the known limit \(\lim_{y \to 0} f(y) = 0\). So we'd like to infer that \(\lim_{x \rightarrow 0} f(g(x)) = 0\).

  But unfortunately \(f\) is not continuous at the point \(0\), and in fact there is no punctured neighborhood around \(0\) where \(g(x) \neq 0\), which would be a required condition for the theorem of the limit of a composite function. So we cannot use that theorem.

  Actually the function is undefined for \(x = \frac{1}{2k\pi}\) , so the limit cannot exist.

• Result

  The limit does not exist.