A very simple question about calculus for middle school students

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A very simple question about calculus for middle school students

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Craftybutterfly

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Craftybutterfly

#1 h Apr 9, 2025, 7:41 AM

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$\lim_{x \to 8} \frac{2x^2+13x+6}{x^2+14x+48}=$ ? (there is an easy workaround)
(I know this is very easy- a little child can solve this in 1 second kinda problem so don't argue or mock me please)

This post has been edited 2 times. Last edited by Craftybutterfly, Apr 10, 2025, 9:26 PM

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#2 h Apr 9, 2025, 7:44 AM

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sol

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Yiyj1

#3 h Apr 9, 2025, 7:49 AM

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#4 h Apr 9, 2025, 2:33 PM

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solution

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KAME06

#5 h Apr 9, 2025, 4:10 PM

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We can plug because the function is continuous in $8$ .
Pretty well known that polynomials are continuous, and $2x^2+13x+6$ and $x^2+14x+48$ are continuous on $8$ and different from $0$ when you evaluate them on $8$ .
That implies that $\frac{2x^2+13x+6}{x^2+14x+48}$ is continuous on $8$ , so $\lim_{x \to 8} \frac{2x^2+13x+6}{x^2+14x+48}=\frac{2(8)^2+13(8)+6}{(8)^2+14(8)+48}$ (factorize if u want to do less work).

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#6 h Apr 10, 2025, 5:22 AM

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Craftybutterfly wrote:

Find $\lim_{x \to -6} \frac{2x^2+13x+6}{x^2+14x+48}.$

FTFY

This post has been edited 1 time. Last edited by HacheB2031, Apr 10, 2025, 2:37 PM

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Craftybutterfly

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#7 h Apr 10, 2025, 5:24 AM

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HacheB2031 wrote:

Craftybutterfly wrote:

Find $\lim_{x \to -8} \frac{2x^2+13x+6}{x^2+14x+48}.$

FTFY

?

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#9 h Apr 10, 2025, 6:02 AM

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@sp0rtman00000Click to reveal hidden text

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#10 h Apr 10, 2025, 2:37 PM • 1 Y

Y by LawofCosine

Craftybutterfly wrote:

HacheB2031 wrote:

Craftybutterfly wrote:

Find $\lim_{x \to -6} \frac{2x^2+13x+6}{x^2+14x+48}.$

FTFY

?

I'm sorry, I realized it should probably be as $x\to-6.$ Let me explain why.
When you make limits, you often want them to tend to a value where a function is not defined. Take the limit $\lim_{x\to0}\frac{\sin x}x.$ Clearly, direct evaluation gives $\frac{\sin0}0=\frac00,$ an indeterminate form, but (using methods such as the squeeze theorem) you can show that $\lim_{x\to0}\frac{\sin x}x=1.$ Despite the function not being defined there, it has a well-defined limit. In my [edited] post, notice that as $x\to-6,$ you also have $x^2+14x+48\to0$ and $2x^2+13x+6\to0,$ making this limit not evaluatable with direct evaluation. (The limit may exist, but in the old post where $x\to-8,$ it blows up.) If you have the limit of a rational function, a function where the numerator and denominator are both polynomials, you can cancel common factors to redefine the function where it wasn't defined. Also, note that you can't always use direct evaluation. Let $f(x)=\begin{cases}x^2&x\ne0\\1&x=0\end{cases}.$ Try to evaluate the limit $\lim_{x\to0}f(x).$ If we directly evaluate it, we get that $\lim_{x\to0}f(x)=f(0)=1.$ However, $f(x)$ has a problem: it isn't continuous, or, if you drew its graph, you would need to pick up your pencil from the paper. The actual limit turns out to be $\lim_{x\to0}f(x)=\lim_{x\to0}x^2=0.$ Because $x^2$ is a polynomial, it is continuous; all polynomials are continuous. As well as all rational functions, over their domain. Anywhere a rational function is defined, it is continuous, except at points where its denominator is $0.$ sidenote

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Craftybutterfly

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#11 h Apr 10, 2025, 6:19 PM

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It is $\lim_{x\to8}$ not $\lim_{x\to-6}$

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yaxuan

#12 h Apr 10, 2025, 8:51 PM

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op wrote:

a kindergartener can solve this in 1 second

Bruh

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#13 h Apr 10, 2025, 9:27 PM

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Craftybutterfly wrote:

sol

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#14 h Apr 14, 2025, 2:27 AM

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whats lim again？。。。

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#15 h Apr 14, 2025, 2:40 AM

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a limit (concept in calculus)

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#16 h Apr 14, 2025, 1:11 PM

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Craftybutterfly wrote:

a kindergartener can solve this in 1 second

I think that is a bit exagerated

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#17 h Apr 14, 2025, 1:12 PM

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Craftybutterfly wrote:

a kindergartener can solve this in 1 second

I think that is a bit exagerated

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#18 h Apr 14, 2025, 2:12 PM

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#19 h Apr 15, 2025, 5:24 AM

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Question: what is an example of a with no limit or a limit of 0

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#20 h Apr 15, 2025, 5:29 AM • 1 Y

Y by LawofCosine

Craftybutterfly wrote:

Question: what is an example of a with no limit or a limit of 0

x-> -8 has no limit
x-> -1/2 gives a limit of 0

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Craftybutterfly

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#21 h Apr 15, 2025, 5:35 AM

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Thx @bove

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