Question

If $f(x) = \begin{cases} \dfrac{x^2 \log(\cos x)}{\log(1+x)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$, then at $x=0$, $f(x)$ is

• A. not continuous
• B. continuous but not differentiable
• C. differentiable
• D. not continuous, but differentiable

Hint:

Use series expansion to check differentiability at points with indeterminate forms.

Answer 1

The Correct Option is C

To evaluate the behavior of the function \(f(x)\) at \(x = 0\), we must analyze its continuity and differentiability at this point.

The function is defined as:

\(f(x) = \begin{cases} \dfrac{x^2 \log(\cos x)}{\log(1+x)}, & x \neq 0 \\ 0, & x = 0 \end{cases}\)

Continuity at \(x = 0\)

For \(f(x)\) to be continuous at \(x = 0\), we need \( \lim_{x \to 0} f(x) = f(0) = 0 \). We examine the limit:

\(\lim_{x \to 0} \dfrac{x^2 \log(\cos x)}{\log(1+x)} \)

Applying L'Hôpital's Rule, which is applicable as both numerator and denominator approach zero:

\(\lim_{x \to 0} \dfrac{2x \log(\cos x) + x^2 \left(-\tan(x)\right)}{\dfrac{1}{1+x}}\)

As \(x \to 0\), \(\log(\cos x) \sim -\dfrac{x^2}{2}\) and \(\log(1+x) \sim x - \dfrac{x^2}{2}\). Simplifying further using L'Hôpital's Rule again gives:

\(\lim_{x \to 0} \dfrac{-x^2}{x} = \lim_{x \to 0} -x = 0\)

Thus, \( \lim_{x \to 0} f(x) = 0 = f(0) \), indicating that \(f(x)\) is continuous at \(x = 0\).

Differentiability at \(x = 0\)

We must check if \(f(x)\) is differentiable at \(x=0\). A function \(f\) is differentiable at a point \(c\) if:

\(\lim_{x \to c} \dfrac{f(x) - f(c)}{x-c}\) exists.

For \(x = 0\), since \(f(0) = 0\), we need:

\(\lim_{x \to 0} \dfrac{f(x)}{x} = \lim_{x \to 0} \dfrac{\dfrac{x^2 \log(\cos x)}{\log(1+x)}}{x}\)

\(\lim_{x \to 0} \dfrac{x \log(\cos x)}{\log(1+x)}\)

Upon simplifying using earlier expansions, it is seen as a standard form where:

\(\sim \dfrac{- x}{x} = -1\)

The derivative exists, so \(f(x)\) is differentiable at \(x = 0\).

Therefore, the correct classification of \(f(x)\) at \(x = 0\) is that it is differentiable.