[SPLIT] three 'find the limit, if it exists' exercises

[fraiha]

Sorry to ask for help so soon but I'm having problems with a number of questions.

1) Find the limit, if it exists. lim (x^2-16) / x-2 as x -> 2
I believe this is Undetermined as I wasn't able to figure out using algebra how to get rid of the denominator and plugging in 2 for x has 0 as the denominator.

2) Prove that lim sqrt(x) e^sin(pi/x) = 0 as x->0+
This one I made a chart of x | f(x), plugged in .1, .01, .001 etc. to show that as X is closer to 0, so is f(x). Is there a more precise way to do this through calculations such as the one above if it were possible?

3) Find the limit, if it exists. lim 1/x - 1|x| as x -> 0+
Well 1/x - 1/x is 1-1/x is 1-1/0 so the answer is 1. But the problem is 1/x + 1/x is 2/0. That means the answer is Undeterminded?

I really am thankful for the help you guys give and I apologize for spamming the boards with my questions.

 

[pka]

For 1), you need to note that \(\displaystyle \L
x^4 - 16 = \left( {x - 2} \right)\left( {x + 2} \right)\left( {x^2 + 4} \right)\) Divide out \(\displaystyle \L \left( {x - 2} \right)\).

For 2), \(\displaystyle \L
x \to 0^ + ,\quad - 1 \le \sin (\pi /x) \le 1\quad \Rightarrow \quad \frac{{ \sqrt x }}{e} \le \sqrt x e^{\sin (\pi /x)} \le \sqrt x e\)

 

[galactus]

1) Find the limit, if it exists. lim (x^2-16) / x-2 as x -> 2
I believe this is Undetermined as I wasn't able to figure out using algebra how to get rid of the denominator and plugging in 2 for x has 0 as the denominator.

There is a vertical asymptote at x=2. The one-sided limits approach \(\displaystyle {+\infty}\)from the left and\(\displaystyle {-\infty}\)from the right.

What does that mean?. Undefined?.

2) Prove that lim sqrt(x) e^sin(pi/x) = 0 as x->0+
This one I made a chart of x | f(x), plugged in .1, .01, .001 etc. to show that as X is closer to 0, so is f(x). Is there a more precise way to do this through calculations such as the one above if it were possible?

You could possibly look at it this way:

\(\displaystyle \L\\\lim_{x\to\0^{+}}\sqrt{x}e^{sin(\frac{\pi}{x})}\)

\(\displaystyle \L\\\lim_{x\to\0^{+}}\sqrt{x}\lim_{x\to\0^{+}}e^{sin(\frac{\pi}{x})}\)

\(\displaystyle \L\\\sqrt{\lim_{x\to\0^{+}}(x)}\lim_{x\to\0^{+}}e^{sin(\frac{\pi}{x})}=0\)

3) Find the limit, if it exists. lim 1/x - 1|x| as x -> 0+
Well 1/x - 1/x is 1-1/x is 1-1/0 so the answer is 1. But the problem is 1/x + 1/x is 2/0. That means the answer is Undeterminded?

\(\displaystyle 1-\frac{1}{0}\) is not 1, it's undefined.

Looking at the one-sided limits, we see they have different values as x approaches 0. What does this mean?. Undefined?.

 

[pka]

There is a point we need to make with \(\displaystyle \L\\\sqrt{\lim_{x\to\0^{+}}(x)}\lim_{x\to\0^{+}}e^{sin(\frac{\pi}{x})}=0\).

The function \(\displaystyle \L\\\sin(\frac{\pi}{x})\) oscillates wildly near 0.

Therefore, it is clear that \(\displaystyle \L\\\lim_{x\to\0^{+}}e^{sin(\frac{\pi}{x})}\not=0\).

So, we must note that \(\displaystyle \L\\\ e^{sin(\frac{\pi}{x})}\) is bounded function!

 

[galactus]

Good point, pka. It's a wacky graph.