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Exercise 5.1 · Q31

Q.Show that the function defined by f(x)=cos⁡(x2)f(x) = \cos (x^2) is a continuous function.

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The function f(x)=cos⁡(x2)f(x) = \cos(x^2) is continuous for all real xx because it is the composition of two continuous functions: g(x)=x2g(x) = x^2 (a polynomial, continuous everywhere) and h(u)=cos⁡uh(u) = \cos u (a trigonometric function, continuous everywhere). By the theorem on continuity of compositions, f=h∘gf = h \circ g is continuous on R\mathbb{R}.

The key idea here is that continuity is preserved under composition. If you have two functions that are each continuous at the relevant points, then their composition is also continuous. This is one of the most powerful shortcuts in analysis — it saves you from having to wrestle with ϵ\epsilon-δ\delta proofs every time.

Let’s break it down.

  1. Identify the structure of ff.

    Notice that f(x)=cos⁡(x2)f(x) = \cos(x^2) can be seen as: first square the input xx, then take the cosine of the result. So define:

    • g(x)=x2g(x) = x^2, which maps R\mathbb{R} to [0,∞)[0, \infty).
    • h(u)=cos⁡uh(u) = \cos u, which maps R\mathbb{R} to [−1,1][-1, 1]. Then f(x)=h(g(x))f(x) = h(g(x)).
  2. Check continuity of the inner function gg.

    g(x)=x2g(x) = x^2 is a polynomial. Every polynomial is continuous at every real number. So gg is continuous on R\mathbb{R}.

  3. Check continuity of the outer function hh.

    h(u)=cos⁡uh(u) = \cos u is a standard trigonometric function. From the definition of cosine (or from its graph), we know cos⁡u\cos u is continuous for all real uu. So hh is continuous on R\mathbb{R}.

  4. Apply the Composition Theorem.

    The theorem states: If gg is continuous at x=ax = a and hh is continuous at g(a)g(a), then the composite function h∘gh \circ g is continuous at x=ax = a.

    Since gg is continuous everywhere and hh is continuous everywhere, the composition is continuous at every a∈Ra \in \mathbb{R}. …

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