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attashe74 [19]
2 years ago
12

Can you also explain how you got the answer

Mathematics
2 answers:
balandron [24]2 years ago
8 0

Answer:

The answer is D

Step-by-step explanation:

The y values are going up by 2

For linear equations you follow y = mx + b

M = 2

B would be 1 since that is what y is when x is 0

So it must be D

Alja [10]2 years ago
5 0

Answer:

I recoment using the formula because its different ways its taught

Step-by-step explanation:

maybe look up different formula's

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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
grandymaker [24]
(2x+1)^{\cot x}=\exp\left(\ln(2x+1)^{\cot x}\right)=\exp\left(\cot x\ln(2x+1)\right)=\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)

where \exp(x)\equiv e^x.

By continuity of e^x, you have

\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)

As x\to0^+ in the numerator, you approach \ln1=0; in the denominator, you approach \tan0=0. So you have an indeterminate form \dfrac00. Provided the limit indeed exists, L'Hopital's rule can be used.

\displaystyle\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\frac2{2x+1}}{\sec^2x}\right)

Now the numerator approaches \dfrac21=2, while the denominator approaches \sec^20=1, suggesting the limit above is 2. This means

\displaystyle\lim_{x\to0^+}(2x+1)^{\cot x}=\exp(2)=e^2
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3 years ago
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