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andrew11 [14]
3 years ago
12

Suppose that ƒ is an odd function of x. Does knowing that limxs0+ ƒ(x) = 3 tell you anything about limxs0- ƒ(x)? Give reasons fo

r your answer
Mathematics
1 answer:
bonufazy [111]3 years ago
6 0

we are given

f(x) is a odd function

so, f(-x)=-f(x)

-f(x)=f(-x)

now, we are given

\lim _{x\to 0+}f\left(x\right)=3

Let's multiply both sides by -1

-\lim _{x\to 0+}f\left(x\right)=-3

\lim _{x\to 0+}-f\left(x\right)=-3

we have

-f(x)=f(-x)

\lim _{x\to 0+}f\left(-x\right)=-3

now, let's consider y=-x

so, when x--->0+

but for y=-x

y--->0-

now, we can plug it

\lim _{y\to 0-}f\left(y\right)=-3

replace y as x

we get

\lim _{x\to 0-}f\left(x\right)=-3.................Answer

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Answer:

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There are 1920 fence post used in a 12 kilometer stretch of fence how many fence post are used in 1 kilometer of fence
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Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

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