Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 2)
This image has been Flagged as inappropriate Click to unflag
Image (2 of 2)
justaguide | Certified Educator

You are not allowed to ask multiple questions. This is a response for 3 of the several that you have posted.

# If `f(x) = sqrt(x+2)` , determine whether f is continuous from the left, from the right or neither for x = -2

`lim_(x->-2^+) f(x)`

= `lim_(x->-2^+) sqrt(x+2)`

= `sqrt(-2+2)`

= 0

`lim_(x->-2^-) f(x)`

= `lim_(x->-2^-) sqrt(x+2)`

But for x < -2, `sqrt(x+2)` is not real as x + 2 is negative and the square root of a negative number is not real.

Therefore `lim_(x->-2^-) f(x) != f(-2)` .

The function is continuous from the right at x = -2.

# If f and g are continuous with f(3) = 5 and `lim_(x->3)[2*f(x) - g(x)] = 4` , find g(3).

Both the functions f and g are continuous at x = 3.

` lim_(x->3)[2*f(x) - g(x)] = 4`

`=> [2*f(3) - g(3)] = 4`

`=> 2*5 - g(3) = 4`

`=> 10 - g(3) = 4`

`=> g(3) = 10 - 4`

`=> g(3) = 6`

# Use the definition of continuity and the properties of limits to show that the function `f(x) = (x+1)/(2x^2 - 1)` is continuous at x = 4

The value of f(x) at x = 4 is `(4+1)/(2*4^2 - 1) = 5/31`

`lim_(x->4^-)(x+1)/(2x^2 - 1)`

= `5/31`

`lim_(x->4^+)(x+1)/(2x^2 - 1)`

= `5/31`

As for a = 4, `lim_(x->a^-) = lim_(x->a^+) = f(a)` , the function is continuous at x = 4.