Let f(x) = \ \beginmatrix ( x - 1 )^1 \over 2 - x, & x \gt 1,x e 2 k\,\,\,\,\,\,\,\,\,\,\,\,\,\, & ,x = 2 \endmatrix . Thevaue of k for which f s continuous at x = 2 is :

Limits Continuity and Differentiability PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q43 with Solution

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Correct answer: $K=\DFRAC{1}{E}$, THE STORED ANSWER $E-1$ APPEARS INCORRECT.

For continuity at $x=2$ , we need $k=\lim_{x\to 2}(x-1)^{\frac{1}{2-x}}.$
This is of the indeterminate form $1^\infty$ , so take logarithm. Let $L=\lim_{x\to 2}(x-1)^{\frac{1}{2-x}}.$ Then $\ln L=\lim_{x\to 2}\frac{\ln(x-1)}{2-x}.$
Put $t=x-1$ . Then as $x\to 2$ , $t\to 1$ . So $\ln L=\lim_{t\to 1}\frac{\ln t}{1-t}.$ This is $0/0$ , so apply L'Hospital's Rule: $\ln L=\lim_{t\to 1}\frac{1/t}{-1}=-1.$ Hence $L=e^{-1}=\frac{1}{e}.$
Therefore, for continuity at $x=2$ , $k=\frac{1}{e}.$
Checking options:

So the correct value is $\boxed{\frac{1}{e}}$ , which is not present in the options.