Lim_(x->1) (x/(x-1)-1/logx) এর মান কত?

প্রশ্ন: \( \lim_{x \to 1} \left( \frac{x}{x-1} - \frac{1}{\ln x} \right) \) এর মান কত?

উত্তর: \( \frac{1}{2} \)

সমাধান:

ধরি, \( x = 1 + h \), যেখানে \( h \to 0 \) যখন \( x \to 1 \)।

তাহলে,

\( \lim_{x \to 1} \left( \frac{x}{x-1} - \frac{1}{\ln x} \right) = \lim_{h \to 0} \left( \frac{1+h}{h} - \frac{1}{\ln(1+h)} \right) \)

আমরা জানি, \( \ln(1+h) = h - \frac{h^2}{2} + \frac{h^3}{3} - \frac{h^4}{4} + \dots \) যখন \( |h| < 1 \)।

সুতরাং,

\( \lim_{h \to 0} \left( \frac{1+h}{h} - \frac{1}{h - \frac{h^2}{2} + \frac{h^3}{3} - \dots} \right) \)

\( = \lim_{h \to 0} \left( \frac{1+h}{h} - \frac{1}{h(1 - \frac{h}{2} + \frac{h^2}{3} - \dots)} \right) \)

\( = \lim_{h \to 0} \left( \frac{1+h}{h} - \frac{1}{h} (1 - \frac{h}{2} + \frac{h^2}{3} - \dots)^{-1} \right) \)

আমরা জানি, \( (1+x)^{-1} = 1 - x + x^2 - x^3 + \dots \) যখন \( |x| < 1 \)।

\( = \lim_{h \to 0} \left( \frac{1+h}{h} - \frac{1}{h} \left( 1 + \left( \frac{h}{2} - \frac{h^2}{3} + \dots \right) + \left( \frac{h}{2} - \frac{h^2}{3} + \dots \right)^2 + \dots \right) \right) \)

\( = \lim_{h \to 0} \left( \frac{1+h}{h} - \frac{1}{h} \left( 1 + \frac{h}{2} + O(h^2) \right) \right) \)

\( = \lim_{h \to 0} \left( \frac{1+h}{h} - \frac{1}{h} - \frac{1}{2} + O(h) \right) \)

\( = \lim_{h \to 0} \left( \frac{1+h-1}{h} - \frac{1}{2} + O(h) \right) \)

\( = \lim_{h \to 0} \left( \frac{h}{h} - \frac{1}{2} + O(h) \right) \)

\( = \lim_{h \to 0} \left( 1 - \frac{1}{2} + O(h) \right) \)

\( = 1 - \frac{1}{2} = \frac{1}{2} \)

সুতরাং, \( \lim_{x \to 1} \left( \frac{x}{x-1} - \frac{1}{\ln x} \right) = \frac{1}{2} \).

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