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Determine the value of the constant k so that the function\(f(x) = \begin{cases} \frac{x^2-3x+2}{x-1}&, \quad if\, x ≠1\\k &, \quad if\, x=1 \end{cases} \)  is continuous at x = 1.

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Meaning of continuity of function – 

If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below :

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If :–

\(\lim\limits_{h \to 0} f(c-h)\) \(\lim\limits_{h \to 0} f(c+h)\) = f(c) .... equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL {i.e, \(\lim\limits_{h \to 0} f(c-h)\)} must be equal to limiting value of right neighbourhood of x = c called right hand limit RHL  {i.e, \(\lim\limits_{h \to 0} f(c+h)\)} and both must be equal to the value of f(x) at x = c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous 

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous. 

Given,

\(f(x) = \begin{cases} \frac{x^2-3x+2}{x-1}&, \quad if\, x ≠1\\k &, \quad if\, x=1 \end{cases} \)….equation 2

We need to find the value of k such that f(x) is continuous at x = 1 

Since f(x) is continuous at x = 1 

∴ (LHL as x tends to 1) = (RHL as x tends to 1) = f(1)

As, 

f(1) = k [from equation 2] 

We can find either LHL or RHL to equate with f(1) 

Let’s find RHL,you can find LHL also.

As, 

f(x) is continuous 

∴ RHL = f(1) 

∴ k = –1

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