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Lesechka [4]
2 years ago
14

I will marke brainliest. what is If 1=3

Mathematics
2 answers:
pogonyaev2 years ago
3 0

Answer:

  • 3

Step-by-step explanation:

<u>Count the number of letters</u>

  • 1 = one ⇒ 3 letters
  • 2 = two ⇒ 3 letters
  • 3 = three ⇒ 5 letters
  • 4 = four ⇒ 4 letters
  • 5 = five ⇒ 4 letters
  • 6 = six ⇒ 3 letters
GalinKa [24]2 years ago
3 0

Change integers to strings and put them

  • 1=one=3
  • 2=two=3
  • 3=three=5
  • 4=four=4
  • 5=five=3
  • 6=six=3
  • 7=seven=5

Hence answer is 3

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hram777 [196]

If you know that

e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving e.

For starters, we have

\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}

Let y=\dfrac34(x+1). Then as x\to\infty, we also have y\to\infty, and

2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2

So in terms of y, the limit is equivalent to

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Now use some of the properties of limits: the above is the same as

\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}

The first limit is trivial; \dfrac1y\to0, so its value is 1. The second limit comes out to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}

To see why this is the case, replace y=-z, so that z\to-\infty as y\to\infty, and

\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e

Then the limit we're talking about has a value of

\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}

# # #

Another way to do this without knowing the definition of e as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)

(where the notation means \exp(x)=e^x, just to get everything on one line).

Recall that

\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)

if f is continuous at x=c. \exp(x) is continuous everywhere, so we have

\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)

For the remaining limit, write

\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}

Now as x\to\infty, both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to

\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83

and our original limit comes out to the same value as before, \exp\left(-\frac83\right)=\boxed{e^{-8/3}}.

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3 years ago
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