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October 26th, 2017, 02:42 PM   #1
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Limits questions help

Hi i'm kinda having problems with these two questions. I wasn't at the lecture and i don't know how to solve them.
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October 26th, 2017, 04:20 PM   #2
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(15) Notice that the denominator, x- 2, does NOT go to 0. So $\displaystyle \lim_{x\to 4}\frac{f(x)- 5}{x-2}= \frac{\lim_{x\to 4} f(x)- \lim_{x\to 4} 5}{\lim_{x\to 4}x- \lim_{x\to 4} 2}= 1$.

I presume you know that $\displaystyle \lim_{x\to 4} x= 4$, that $\displaystyle \lim_{x\to 4}5= 5$, and that $\displaystyle \lim_{x\to 4} 2= 2$.

(16a) Well, basically, you do what the problem tells you to do! Since you are asked to assure that $\displaystyle |f(x)- L< \epsilon$, start by calculating [math]|f(x)- L[/math[ for each problem. In the first problem, $\displaystyle f(x)= \sqrt{x}= x^{1/2}$, $\displaystyle L= 1/2$, and $\displaystyle \epsilon= 1/10$ so you need to look at $\displaystyle |f(x)- L|= |\sqrt{x}- 1/2|< 1/10$. Removing the absolute value, $\displaystyle -1/10< \sqrt{x}- 1/2< 1/10$, $\displaystyle -1/10+ 1/2= 2/5< \sqrt{x}< 1/10+ 1/2= 3/5$.

Now we want to look at $\displaystyle |x- x_0|= |x- 1/4|$.

Squaring the previous inequality $\displaystyle 0\le x< 9/24$ so $\displaystyle x- 1/4< 9/24- 6/24= 3/24$. Each step is "reversible" so if $\displaystyle |x- 1/4|< 3/24$, it will be true that $\displaystyle |\sqrt{x}- 1/2|< 1/10$.

Do b and c in the same way.

Last edited by Country Boy; October 26th, 2017 at 04:50 PM.
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October 26th, 2017, 11:48 PM   #3
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I can't understand why the squaring of inequality is 0≤x<9/24
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October 27th, 2017, 04:57 AM   #4
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Since the square root of x is a real number, x cannot be negative. Since the square root of x is less than 3/5 x cannot be larger than 9/25.

The "24" is a typo. The final result should be 9/25- 1/4= 35/100- 25/100= 10/100= 1/10 so |x- 1/4|< 1/10.

Sorry about that.

Last edited by Country Boy; October 27th, 2017 at 05:01 AM.
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