Squeeze Theorem Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

easy
Find lim⁡x→0xcos⁡ ⁣(1x)\displaystyle\lim_{x \to 0} x\cos\!\left(\frac{1}{x}\right).

Solution

  1. 1
    −1≤cos⁡(1/x)≤1-1 \leq \cos(1/x) \leq 1, so −∣x∣≤xcos⁡(1/x)≤∣x∣-|x| \leq x\cos(1/x) \leq |x|.
  2. 2
    lim⁡x→0(−∣x∣)=0\lim_{x\to0}(-|x|)=0 and lim⁡x→0∣x∣=0\lim_{x\to0}|x|=0.
  3. 3
    By squeeze theorem: limit =0= 0.

Answer

00
Same technique: bound the oscillating factor between -1 and 1, multiply by the vanishing factor, apply squeeze theorem.

About Squeeze Theorem

If g(x)≤f(x)≤h(x)g(x) \leq f(x) \leq h(x) near x=ax = a, and lim⁡x→ag(x)=lim⁡x→ah(x)=L\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, then lim⁡x→af(x)=L\lim_{x \to a} f(x) = L.

Learn more about Squeeze Theorem →

More Squeeze Theorem Examples

Example 1 easy

Use the squeeze theorem to find [formula].

Example 2 hard

Use the squeeze theorem to prove [formula].

Example 4 medium

Show that [formula].

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