Limits, Continuity & Differentiability Question - None

Question

Paragraph for Questions 630 and 631
Let $f: R \to R$ is a function defined by $f(x) = \begin{cases} 1, & \text{if } x = 1 \ e^{(x^{10}-1)} + (x-1)^2 \sin\left(\dfrac{1}{x-1}\right), & \text{if } x \neq 1 \end{cases}$

The value of $f'(1)$ is:

Accepted Answer

Step 1: Verify continuity at x=1

For x ≠ 1: f(x) = e^(x^10-1) + (x-1)^2 sin(1/(x-1))

As x → 1: e^(x^10-1) → e^0 = 1

Since |(x-1)^2 sin(1/(x-1))| ≤ (x-1)^2 → 0, we have lim(x→1) f(x) = 1 = f(1)

So f is continuous at x=1.

Step 2: Apply definition of derivative

f'(1) = lim(h→0) [f(1+h) - f(1)]/h

= lim(h→0) [e^((1+h)^10-1) + h^2 sin(1/h) - 1]/h

Step 3: Expand e^((1+h)^10-1)

(1+h)^10 - 1 = 10h + 45h^2 + ... (by binomial theorem)

e^(10h + 45h^2 + ...) = 1 + (10h + 45h^2 + ...) + O(h^2)

= 1 + 10h + O(h^2)

Step 4: Evaluate the limit

f'(1) = lim(h→0) [10h + h^2 sin(1/h) + O(h^2)]/h

= lim(h→0) [10 + h sin(1/h) + O(h)]

= 10 + 0 = 10

∴ Answer: C (f'(1) = 10)

Question:

Step-by-Step Solution

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