341@RPF, c is dependent on x.
Suppose f be a continuously differentiable function on [a,b] and twice differentiable at x=a with f''(a) being non-zero; that is, the limit
\lim_{x\to a^+}\dfrac{f'(x)-f'(a)}{x-a}
exists and is non-zero.
Applying LMVT to f in the interval [a,x] for any x≤b, we get that exists some c in (a,x) such that
f'(c) = f(x)-f(a)x-a.
Question: Find the following limit:
\lim_{x\to a^+}\dfrac{c-a}{x-a}
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8 Answers
341Hope this is right:
\lim_{x \rightarrow a} \frac{c-a}{x-a} = \lim_{x \rightarrow a} \frac{c-a}{f'(c) -f'(a) } \frac{f'(c)-f'(a)}{x-a}
Now we claim that Lt x→ a \frac{f'(c)-f'(a)}{c-a}= f"(a)
Proof: We know that \lim_{x \rightarrow a^{+}} \frac{f'(x)-f'(a)}{x-a}= f"(a)
This implies that whenever x-a < \delta we have
\left|\frac{f'(x) - f'(a)}{x-a} - f"(a) \right| < \epsilon
Since c-a < x-a < \delta we also have
\left|\frac{f'(c) - f'(a)}{c-a} - f"(a) \right| < \epsilon and hence we have proved that the required limit is f"(a).
Now we tackle \lim_{x \rightarrow a^+} \frac{f'(c) - f'(a)}{x-a}
Remember that f'(c) = \frac{f(x) - f(a)}{x-a}
Hence the given limit may be rewritten as
\lim_{x \rightarrow a^+} \frac{ \frac{f(x) - f(a)}{x-a} - f'(a)}{x-a} = \lim_{x \rightarrow a^+} \frac{ (f(x) - f(a)) - f'(a)(x-a)}{(x-a)^2}
By continuity of f(x), the limit is of the 0/0 form and so we employ L'Hospital's rule to get the equivalent limit
\lim_{x \rightarrow a^+} \frac{ f'(x) - f'(a)}{2(x-a)} = \frac{f"(a)}{2}
Thus the given limit evaluates to \frac{1}{f"(a)} \times \frac{f"(a)}{2} = \frac{1}{2}
66Sorry for the delayed response.
And indeed the answer is 1/2. And the method is the same as prophet sir.