With an example, calculation of limits for a piecewise function is discussed. The conditions under which a function is defined are illustrated with examples.

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Given piecewise function

`f(x)={(1, quad text(for)quad x>=1),(0.5, quad text(for)quad x<1):}`

What is `f(x)|_(x=1)`?

- `0.5`
- `-1`
- `+1`
- `+1`

The answer is '`1`', as given by definition of the function.

Given function

`f(x)={(1, quad text(for)quad x>=1),(0.5, quad text(for)quad x<1):}`

what is left-hand-limit `lim_(x->1-)f(x)`?

- `0.5`
- `0.5`
- `-1`
- `+1`

The answer is '`0.5`'

`lim_(x->1-)f(x)`

`quad quad = f(x)|_(x=1-delta)`

`quad quad = 0.5 quad quad quad quad` (as `1-delta<1`)

Given function

`f(x)={(1, quad text(for)quad x>=1),(0.5, quad text(for)quad x<1):}`

what is right-hand-limit `lim_(x->1+)f(x)`?

- `0.5`
- `-1`
- `+1`
- `+1`

The answer is '`1`'

`lim_(x->1+)f(x)`

`quad quad = f(x)|_(x=1+delta)`

`quad quad = 1 quad quad quad quad` (as `1+delta>1`)

Given function

`f(x)={(1, quad text(for)quad x>=1),(0.5, quad text(for)quad x<1):}`

• `f(1)=1`

• `lim_(x->1-)f(x) = 0.5`

• `lim_(x->1+)f(x) = 1`

The function is not continuous at `x=1`. But, the function is defined by the given definition of the function.

If the function evaluates to a real number, then the * function is defined* at that input value. The limit may not be defined.

**Function is defined by value: ** If `f(a) in RR`, and `lim_(x->a-)f(x)` `!= lim_(x->a+)f(x)`, then function is not continuous but defined by the definition of the function.

*Solved Exercise Problem: *

Given function

`f(x)={(2x, quad text(for)quad x>2),(x^2, quad text(for)quad x=2),(|x|+2, quad text(for)quad x<2):}`

Examine the function at `x=2`.

- continuous
- defined by limit
- defined by value
- All the above
- All the above

The answer is 'All the above'.

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