# Homework Help: Should the limit even exist?

1. Jan 18, 2009

### gabby989062

1. The problem statement, all variables and given/known data
Hello, I have some Calculus problems from my latest homework.

1.
The limit as x approaches 3 of sqrt(x+1)
The back of the book says 2. But I'm wondering why it exists as sqrt(3+1) = sqrt(4) = +-2, so since there are 2 values for this, should the limit even exist?

2.
The limit of delta x as it approaches 0 of (sqrt(x + delta x) - sqrt(x)) divided by delta x.

In similar problems, I factored the denominator out of the numerator so I could cancel the denominator so it could never go to 0. But I don't know how to do this for stuff in the radical sign. The book gives no example of this type of problem.

3. Graphical, numerical, and analytical
a. use a graphing utility to graph the function and estimate the limit.
b. use a table to reinforce your conclusion
c. find the limit by analytic methods.

limit as x approaches 1 from the negative side of 2/(x^2-1).

Since it's from the negative side, I just plugged 0.9999999 for x into my calculator, so the bottom approached 0 as a negative number. So I concluded that it is negative infinity. But isn't plugging in 0.999999 considered the numerical method? How do I do this analytically?

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 18, 2009

### jgens

Re: limits

Regarding one: The answer should not be +-2 as the sqrt(x+1) explicitly denotes the positive square root. Hence, if I asked you what sqrt(49) is equivalent to, you likely would not say +-7 but rather 7. However, given x^2 = 49 and asked to solve for x, you should give the answers +-7.

Regarding two: I suggest that you multiply and divide by the conjugate of sqrt(x+deltax) - sqrt(x). sqrt(x+deltax) + sqrt(x) should be the conjugate.

For number three: The only way I can think of doing it somewhat analytically would be to note that for x < 1 your expression is negative; therefore, as x tends to one from x < 1, f(x) -> -infinity

Last edited: Jan 18, 2009
3. Jan 18, 2009

### Mentallic

Re: limits

This is a common misconception that results from a poor understanding of what is happening in, lets say, making x the subject in $$x^2=y$$. Yes it will result in $$x=\pm \sqrt{y}$$ but because if you square either a positive or negative, you result in a positive. However, if you take $$\sqt{x}$$, x can ony be positive, not negative. So the limit here is only +2.

Try multiplying through by the conjugate of the numerator
Remember, $$(x+y)(x-y)=x^2-y^2$$ so the radicals will be removed (in the numerator at least).

I've never been taught how to find limits that tend to infinite by analytical methods, rather I use observational and logical techniques like yourself.
I can't be of much help here, but I can lend a hand.

How about if we tried dividing through by the highest power of x?

$$lim\frac{2}{x^2-1}=lim\frac{\frac{2}{x^2}}{1-\frac{1}{x^2}}=\frac{2}{1-1^{+}}$$
The denominator will tend to 0 from the negative end, thus negative infinite?