2 years ago

Help me with this please please pleasee

Mathematics

2 answers:

Mazyrski [523]2 years ago

5 0

Answer:

37.79924

Step-by-step explanation:

Veronika [31]2 years ago

5 0

37.79924 that will be your answerw

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3 years ago

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

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