Limits and Continuous Functions problem

[adillhoff]

Homework Statement

Define the function at a so as to make it continuous at a.
$$f(x)=\frac{4-x}{2-\sqrt{x}}; a = 4$$

Homework Equations

$$\lim_{x \rightarrow 4} \frac{4-x}{2-\sqrt{x}}$$

The Attempt at a Solution

I cannot think of how to manipulate the denominator to achieve f(4), so I start by finding the left and right hand limits then applying the informal definition of a limit.

$$\lim_{x \rightarrow +4}\frac{4-x}{2-\sqrt{x}} = 4$$
$$\lim_{x \rightarrow -4}\frac{4-x}{2-\sqrt{x}} = 4$$

I do not know if this is the right way to solve this problem though. If I use this method, I am still left with a function where f(4) is discontinuous. Any hints would be greatly appreciated.

 

[eumyang]

Use the difference of two squares pattern
$$a^2 - b^2 = (a - b)(a + b)$$
where a² - b² corresponds to 4 - x. See what happens.

 

[adillhoff]

Wow that seems so obvious now. I completely overlooked it. Thank you so much for the help.

 

[Mark44]

Homework Statement

Define the function at a so as to make it continuous at a.
$$f(x)=\frac{4-x}{2-\sqrt{x}}; a = 4$$

Homework Equations

$$\lim_{x \rightarrow 4} \frac{4-x}{2-\sqrt{x}}$$

The Attempt at a Solution

I cannot think of how to manipulate the denominator to achieve f(4), so I start by finding the left and right hand limits then applying the informal definition of a limit.

$$\lim_{x \rightarrow +4}\frac{4-x}{2-\sqrt{x}} = 4$$
$$\lim_{x \rightarrow -4}\frac{4-x}{2-\sqrt{x}} = 4$$

Your notation needs a bit of tweaking. What you think you wrote, and what you actually wrote are two different things. Your limits were supposed to be one-sided limits around 4. The second limit, above, was as x approaches -4, which is not in the domain of the square root function.

This is what you should have written:
$$\lim_{x \rightarrow 4^+}\frac{4-x}{2-\sqrt{x}} = 4$$
$$\lim_{x \rightarrow 4^-}\frac{4-x}{2-\sqrt{x}} = 4$$

Notice that the plus and minus signs appear after 4.

I do not know if this is the right way to solve this problem though. If I use this method, I am still left with a function where f(4) is discontinuous. Any hints would be greatly appreciated.

 

[kmacinto]

Can you use L'Hospital's rule for something like this? The limit is in the indeterminate form of 0/0 and the result of one iteration of L'Hospital is 2√x which is continuous at x=4... Am I cheating here?

 

[HallsofIvy]

Yes, you could use L'Hopital's rule but that would be "over-kill". eumyang's suggestion is much simpler.

 

[kmacinto]

Wow that seems so obvious now. I completely overlooked it. Thank you so much for the help.

I guess it doesn't seem obvious to me using the difference of squares then... Would you mind writing it out?

Ken

 

[gb7nash]

I guess it doesn't seem obvious to me using the difference of squares then... Would you mind writing it out?

Ken

$$4 - x = (2 - \sqrt{x})(2 + \sqrt{x})$$

After that, you get cancellation.

 

[kmacinto]

Omg! I was working the pants off the denominator and totally missed that the numerator was the key! Thanks!
Ken

 

[HallsofIvy]

Another, similar, way to do this is to multiply both numerator and denominator by $2+ \sqrt{x}$.