Question

If \(f(x)=\frac{\cos^{4}x-1}{x^{2}}\), \(x\ne0\) and \(f(0)=2\) is a real valued function, then:

• A. \(\lim_{x\rightarrow0}f(x)\) does not exist
• B. \(\lim_{x\rightarrow0}f(x)=1\)
• C. f is not continuous at \(x=0\)
• D. f is continuous at \(x=0\)

Hint:

To test for continuity, always verify that the limit matches the function value at that point. If they differ, the function has a removable discontinuity.

Answer 1

The Correct Option is C

Step 1: State the test.
$f$ is continuous at $0$ only if $\lim_{x\to0}f(x)=f(0)$. Here $f(0)=2$, and away from $0$, $f(x)=\dfrac{\cos^4x-1}{x^2}$.
Step 2: Factor the numerator.
$\cos^4x-1=(\cos^2x-1)(\cos^2x+1)$, and $\cos^2x-1=-\sin^2x$. So \[ f(x)=\frac{-\sin^2x\,(\cos^2x+1)}{x^2}. \]
Step 3: Split into known pieces.
\[ f(x)=-\left(\frac{\sin x}{x}\right)^2(\cos^2x+1). \]
Step 4: Use standard limits.
As $x\to0$, $\dfrac{\sin x}{x}\to1$ and $\cos x\to1$, so $\cos^2x+1\to2$.
Step 5: Evaluate the limit.
\[ \lim_{x\to0}f(x)=-(1)^2(2)=-2. \]
Step 6: Compare with $f(0)$.
Since $-2\ne2=f(0)$, the limit does not match the defined value, so $f$ is not continuous at $x=0$.
\[ \boxed{\text{f is not continuous at }x=0} \]