Calc Quiz: Does a Limit Exist as x Approaches 1?

[Joshrk22]

Yesterday, I took a quiz in my Calc class. My teacher threw some limits on there which I believed were easy. One of them was...

$$\lim_{x\rightarrow 1}\frac{1}{(x-1)^2}$$

For this one I put "Limit does not exist as x approaches 1." I come in today and she tells the class that a limit does exist. I didn't want to argue with her but I trusted my gut feeling and believed we (the class) were right. I come home and check my other Calc book and in there it says that a limit does not exist. So what is it? Is she right, or is my class wrong?

 

[d_leet]

It does not exist in the sense that it does not approach any single number, sometimes people will write in cases like this where both the right, and left hand limits increase without bound that the limit is positive infinity.

 

[ace123]

Seems right to me. Maybe your teacher was mistaken about which problem she was talking about.

 

[Joshrk22]

Thanks guys! I'm going to show her the book tomorrow. Also what about this one?

$$\lim_{x\rightarrow 1}\frac{\sqrt{x-3}-2}{(x-1)}$$

I plugged it into my calculator and it doesn't exist either. Yet again, she said that it does...

 

[d_leet]

That's even worse as the square root of -2 is not even a real number, so if we are dealing with only real numbers that function isn't even defined in a neighbourhood about -1.

 

[Joshrk22]

Yep, and she tried to tell us that none of them were "Does Not Exist..." Hah!

 

[TMM]

Thanks guys! I'm going to show her the book tomorrow. Also what about this one?

$$\lim_{x\rightarrow 1}\frac{\sqrt{x-3}-2}{(x-1)}$$

I plugged it into my calculator and it doesn't exist either. Yet again, she said that it does...

Are you sure it isn't $$\sqrt{x+3}$$?

Because then you can apply l'hopital's rule and it'll exist. Otherwise I dunno.

 

[rook_b]

It sounds to me like she is testing you on material she hasn't yet taught you. A "well the book says" argument might work.

 

[ZioX]

Yesterday, I took a quiz in my Calc class. My teacher threw some limits on there which I believed were easy. One of them was...

$$\lim_{x\rightarrow 1}\frac{1}{(x-1)^2}$$

It diverges to infinity. Or converges at infinity. Take your pick.

 

[dynamicsolo]

This one,

$$\lim_{x\rightarrow 1}\frac{1}{(x-1)^2}$$ ,

is an example of an infinite limit. Because the denominator is always positive (except at x = 1), the one-sided limits from either side of x = 1 go to the same signed infinity, so the limit is taken to be positive infinity. (There is a formal definition for infinite limits which can be used to prove this.)

Were the limit

$$\lim_{x\rightarrow 1}\frac{1}{(x-1)}$$ or

$$\lim_{x\rightarrow 1}\frac{1}{(x-1)^3}$$ ,

then the sign of the denominator would depend on which side x were approaching 1 from:
the limit "from below" would be negative infinity and the limit "from above" would be positive infinity. In these cases, because the one-sided limits don't agree, the two-sided limit for either of these examples indeed does not exist.

The other limit is almost certainly intended to be

$$\lim_{x\rightarrow 1}\frac{\sqrt{x+3}-2}{(x-1)}$$ .

Since x = 1 is not in the domain of the function for the expression you posted, you would be correct that the limit doesn't exist (because f(x) is undefined for x <= 3). For this corrected limit, L'Hopital's Rule will surely work, but it's awfully high-powered and actually involves more writing than the simpler method of "conjugate factors".

Multiply numerator and denominator by $$\sqrt{x+3}+2$$. The numerator now contains the factors of a difference of two squares, which multiply out to leave (x + 3) - 4 = x - 1 . This factor now cancels top and bottom, leaving

$$\lim_{x\rightarrow 1}\frac{1}{\sqrt{x+3}+2}$$ ,

which equals 1/4 . (It seems to me this must be what your instructor meant...)