Q.Prove that the function is continuous at , at and at .
The function is a polynomial, and all polynomials are continuous everywhere on . We verify this at each given point by checking that — the limit equals the function value. The function is continuous at , , and .
The idea of continuity at a point is simple: as you zoom in on the graph at that point, there should be no break, jump, or hole. Formally, for a function to be continuous at , three things must hold:
- is defined.
- exists.
- .
For a linear function like , the graph is a straight line with no breaks anywhere. So we already know it's continuous at every real number. But the problem asks us to prove it at three specific points, which means we show the limit condition holds at each.
For a linear function , we have for any real .
Let's do it step by step for each point.
- At First, , so the function is defined. Now compute the limit as approaches 0:
Since , the function is continuous at .
- At . The limit:
Again, limit equals function value, so continuity holds at .
- At . The limit:
So continuity is satisfied at as well.
A common mistake is to think you need to use the - definition for every such problem. For a linear polynomial, direct substitution into the limit is perfectly valid because the limit of a polynomial as is just the polynomial evaluated at . No need for extra machinery.
If you ever forget, remember: polynomials are continuous on all real numbers. So for any polynomial , . This saves time in exams.
Thus, we have proven that is continuous at all three points.
The function is continuous at , , and .
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