If Eggs and Toast cost $8.67, and they are 15% off, how much do they cost now?

Mathematics

2 answers:

Talja [164]3 years ago

6 0

The eggs and toast cost $7.37

timofeeve [1]3 years ago

3 0

In case you are asked how you came up with your answer..
If you take $8.67 times 0.15 = 1.30
$8.67 minus $1.30 = $7.37

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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

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