Reviewing a Failed Assignment: Evaluating the Limit

[ladyrae]

I’m reviewing an assignment that didn’t go so well

Evaluate limit

lim x->-1 (1/2+(x/(x+3))/(x+1)

I reduced to (1+2x)/2 = -1/2

I originally had x-1/x=2 but i redid the problem as above

Is this correct?

 

[master_coda]

No, the answer is 3/4.

\frac{0.5+\frac{x}{x+3}}{x+1}

You want to multiply both the numerator and denominator by x+3 so that you can get rid of the fraction in the numerator. Then it should be easy to factor x+1 out of both the numerator and denominator.

 

[TALewis]

That's not the correct answer. You want to find:

<br /> \lim_{x\rightarrow -1} \frac{\frac{1}{2} + \frac{x}{x+3}}{x+1}<br />

Move the (x+1) up into the numerator:

<br /> \frac{1}{2(x+1)} + \frac{x}{(x+3)(x+1)}<br />

Get a common denominator and add these two terms together, and you should see that the x+1 will fall out of the top and bottom.

Edit: master_coda's way of multiplying the top and bottom by (x+3) is probably simpler.

You should get 3/4, like master_coda said.

Like I said in another of your threads, you can check your work by plugging in a value close to -1 into your original expression (like -0.99999). Your numerical result should be close to the analytical answer. If they aren't, you've probably made a mistake.

 

[ladyrae]

Notation

I'm a newbie ...

Are you using software to write the equation in that format?

 

[TALewis]

It's LaTeX markup (code), built into these forums. Check out this thread:

https://www.physicsforums.com/showthread.php?t=8997

The best way to learn is by example. You can click on any equation in any thread and see what the author typed to create it.