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प्रश्न
Prove that f(x) = 2x2 + 3x - 5 is continuous at all points in R
योग
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उत्तर
f(x) = 2x2 + 3x – 5
Clearly f(x) is defined for all points of R.
Let x0 be an arbitrary point in R.
Then f(x0) = 2x02 + 3x0 – 5 .......(1)
`lim_(x -> x_0) f(x) = lim_(x -> x_0) (2x^2 + 3x - 5)`
= 2x02 + 3x0 – 5 .......(2)
From equation (1) and (2)
`lim_(x -> x_0) f(x) = f(x_0)`
Thus, f(x) is defined at all points of R limit of f(x) exist at all points of R and is equal to the value of the function f (x).
Thus f(x) is continuous at all points of R.
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