Use Continuity to evaluate the limit

[kssthestrs]

Consider the intervals for which the numerator and the denominator are continuous.

lim x→4 12 + sqrtx all over sqrt12 + x

The numerator 12 + sqrt x is continuous on the interval.


For this part of the question I got [0, +infinity)

The denominator sqrt12 + x is continuous and nonzero on the interval This second part I am not sure about. Is it (-infinity, +infinity) ?? Can someone help me out please?

 

[Quaid]

Consider the intervals for which the numerator and the denominator are continuous.

lim x→4 12 + sqrtx all over sqrt12 + x

The numerator 12 + sqrt x is continuous on the interval.


For this part of the question I got [0, +infinity)

The denominator sqrt12 + x is continuous and nonzero on the interval This second part I am not sure about. Is it (-infinity, +infinity) ?? Can someone help me out please?

Hi kssthestrs:

The denominator is continuous on (-∞, ∞) but it is not nonzero for all Real numbers x.

:idea: The graph of y = sqrt(12) + x is a straight line with slope 1, so it must have an x-intercept.

Solve the equation 0 = sqrt(12) + x, to find the x-intercept. Use that value to correct your interval notation.


f(x) = [12 + sqrt(x)]/[sqrt(12) + x]

The limit concerns the behavior of f, as x approaches 4. Are both the numerator and denominator continuous at x=4? If they are, then f(x) is continuous at x=4, and you may use the definition of continuity to say that the limit equals f(4).

Cheers

 

[kssthestrs]

Thank you!! I appreciate the help.

 

[Quaid]

You're welcome. Did you notice my typographical error? (Now corrected.)

I had mixed up your exercise with somebody else's, when I wrote that your limit equals f(1). I should have typed "x approaches 4" and f(4), instead.