Limits, Continuity & Differentiability Question - None

Question

Let $f(x) = \begin{cases} \left[1 + \ln(c^2 + c + 1)\tan^2(x-1)\right]^{\frac{1}{(\ln x)^2}}, & x \neq 1 \ 3c, & x = 1 \end{cases}$ where $c \in R$. If $\lim_{x \to 1} f(x)$ exists but $f(x)$ is discontinuous at $x = 1$, then $c$ can take the value:

Accepted Answer

Step 1: Identify the form as x→1

As x→1: tan²(x-1)→0, so the base [1 + ln(c²+c+1)tan²(x-1)]→1

The exponent 1/(ln x)² → ∞ as x→1

This gives the indeterminate form 1^∞

Step 2: Apply standard limit technique

For lim(x→a) [1 + g(x)]^(1/h(x)) = 1^∞ to exist and be finite: lim(x→a) g(x)/h(x) must exist.

Here: lim(x→1) [ln(c²+c+1)tan²(x-1)] · (ln x)² = lim(x→1) ln(c²+c+1) · tan²(x-1) · (ln x)²

Step 3: Evaluate using Taylor series

As x→1: tan(x-1) ≈ (x-1), so tan²(x-1) ≈ (x-1)²

Also: ln x = ln(1+(x-1)) ≈ (x-1), so (ln x)² ≈ (x-1)²

Thus: ln(c²+c+1) · (x-1)² · (x-1)² → 0 if c²+c+1 > 0 (always true)

Step 4: Conclude for discontinuity

The limit = e^0 = 1

For discontinuity: f(1) = 3c ≠ 1

Therefore: 3c ≠ 1, so c ≠ 1/3

Also need c²+c+1 > 0 (always satisfied)

∴ Answer: A (c can be any real value except c = 1/3)

Question:

Step-by-Step Solution

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