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Xelga [282]
3 years ago
12

What is lim x → 0 e^2x - 1/ e^x - 1

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Hello, please consider the following.

\displaystyle \forall x \in \mathbb{R}\\\\\dfrac{e^{2x}-1}{e^x-1}\\\\=\dfrac{(e^x)^2-1^2}{e^x-1}\\\\=\dfrac{(e^x-1)(e^x+1)}{e^x-1}\\\\=e^x+1\\\\\text{So, we can find the limit.}\\\\\lim_{x\rightarrow 0} \ {\dfrac{e^{2x}-1}{e^x-1}}\\\\=\lim_{x\rightarrow 0} \ {e^x+1}\\\\=e^0+1\\\\\large \boxed{\sf \bf \ =2 \ }

Thank you

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djyliett [7]

Step-by-step explanation:

Hey, there!!

Let's simply work with it,

60° + 2x° + (2x+12)° = 180° { being a linear pair}.

or, 72° + 4x = 180°

4x = 180° - 70°

x =  \frac{108}{4}

Therefore, the value of x is 27°.

Now,

2x° = (2×27)°= 54°

(2x+12)° = 2×27°+12°= 66°.

(x+2)= 27+ 2= 29°

Now, The left angle is 27°.

Therefore, The answer is option A.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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3 years ago
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kykrilka [37]

Answer:

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2 years ago
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erica [24]

Answer:

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WARRIOR [948]

Answer:

-1 1/3

Step-by-step explanation:

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-4/3

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4 0
2 years ago
Read 2 more answers
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