Question

Let $\begin{align}
  & f(x)={{(x-1)}^{\dfrac{1}{2-x}}},x>1,x\ne 2 \\
 & =k,x=2
\end{align}$ . The value of k for which f is continuous at x = 2 is
( a ) ${{e}^{-2}}$
( b ) $e$
( c ) ${{e}^{-1}}$
( d ) 1

Answer 1

Hint: To solve this question first we will check which type of indeterminate form we get on direct substitution of x = 2, then we will use the method to solve the indeterminate form ${{1}^{\infty }}$case, by $\displaystyle \lim_{x \to \infty }{{\left( f(x) \right)}^{g(x)}}={{e}^{\displaystyle \lim_{x \to \infty }g(x)[f(x)-1]}}$.

Complete step-by-step answer:
Before we solve this question, let's see what the limit of a function means.
Now, we know that a function f ( x ) assigns an output y = f ( x ) for every input x. we say limit of a function is L as x moves closer to p that is, if f ( x ) gets closer and closer to L when f is applied to any input sufficiently close to input p, then limit of function f ( x ) is L at point P.
If the expression obtained after substitution does not provide sufficient information to determine the original limit, then it is said to be an indeterminate form.
Now, in question it given that at x > 1 and $x\ne 2$, function is defined as $f(x)={{(x-1)}^{\dfrac{1}{2-x}}}$and at x = 2, we have f ( x ) = k and it is asked to determine the value of k.
Now, at $x\ne 2$, function is defined as $f(x)={{(x-1)}^{\dfrac{1}{2-x}}}$ and at x = 2, f ( x ) = k
For existence of limit, Left hand limit = Right hand limit
So, at x = 2,
$\displaystyle \lim_{x \to 2}{{(x-1)}^{\dfrac{1}{2-x}}}=k$
Now, putting x = 2 in$f(x)={{(x-1)}^{\dfrac{1}{2-x}}}$, we get ${{1}^{\infty }}$ case, which is indeterminate form.
Now, we know that $\displaystyle \lim_{x \to \infty }{{\left( f(x) \right)}^{g(x)}}={{e}^{\displaystyle \lim_{x \to \infty }g(x)[f(x)-1]}}$ , for ${{1}^{\infty }}$case.
So, in ${{(x-1)}^{\dfrac{1}{2-x}}}$ f ( x ) = x – 1 and g ( x ) = $\dfrac{1}{2-x}$
$\displaystyle \lim_{x \to 2}{{(x-1)}^{\dfrac{1}{2-x}}}=k$
${{e}^{\displaystyle \lim_{x \to 2}\dfrac{1}{2-x}[(x-1)-1]}}=k$
${{e}^{\displaystyle \lim_{x \to 2}\dfrac{1}{2-x}\cdot (x-2)}}=k$
On simplifying, we get
${{e}^{\displaystyle \lim_{x \to 2}\dfrac{1}{-(x-2)}\cdot (x-2)}}=k$
\[{{e}^{\displaystyle \lim_{x \to 2}-1}}=k\]
Putting limit, we get
\[{{e}^{-1}}=k\]
So, f is continuous at x = 2 for k = \[{{e}^{-1}}\].

So, the correct answer is “Option c”.

Note: To do such a question one must know the definition of limits and it’s application too. Indeterminate form are important topic, so one must know all type of indeterminate forms of limit such as ${{1}^{\infty }}$, ${{0}^{\infty }}$, $\infty \times \infty $, $0\times 0.$ and always remember that limit only exist if Left hand limit = Right hand limit.