So, Should the limit even exist?

[gabby989062]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[jgens]

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

 

[Mentallic]

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?

This is a common misconception that results from a poor understanding of what is happening in, let's 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.

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.

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).

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?

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?