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Ganezh [65]
2 years ago
11

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

Mathematics
1 answer:
e-lub [12.9K]2 years ago
3 0
sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

\large  \lim_{ x \to 0  } ~  x^x = 1

I will make the assumption that <span>log(x)=ln(x)</span><span>.

The limit result can be proven if the base of </span><span>log(x)</span><span> is 10. 
</span>
\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x  \log x }  \\~\\  \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x)  }   \\~\\  \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x)  }  ~~ \normalsize{\text{ substituting x for sin x } } \\~\\   \large  = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log(  \lim_{x \to 0^{+}}x^x)  } = \frac{1-1}{\log(1)}   = \frac{0}{0}

We get the indeterminate form 0/0, so we have to use <span>Lhopitals rule 

</span>\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x)  } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x)  }   \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1
<span>
Therefore,

</span>\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x  \log x }  =\boxed{ -1}<span>
</span>
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A shoreline observation post is located on a cliff such that the observer is 280 feet above sea level. The observer spots a ship
morpeh [17]

Answer:

Step-by-step explanation:

a) D/280 = tan(90 - 6)

   D = 280tan84 = 2,664 ft

b) D/280 = tan(90 - 16)

   D = 280tan74 = 976 ft

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3 years ago
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Nostrana [21]

Answer:

\displaystyle   \begin{cases} \displaystyle  {x} _{1} =  - p \\   \displaystyle x _{2}   =  -  q \end{cases}

Step-by-step explanation:

we would like to solve the following equation for x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}  +  \frac{1}{x}  =  \frac{1}{p  + q + x}

to do so isolate \frac{1}{x} to right hand side and change its sign which yields:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{1}{p  + q + x}  -  \frac{1}{x}

simplify Substraction:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{x - (q + p +  x)}{x(p  + q + x)}

get rid of only x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{  - (q + p )}{x(p  + q + x)}

simplify addition of the left hand side:

\displaystyle  \frac{q + p}{pq}     =  \frac{  - (q + p )}{x(p  + q + x)}

divide both sides by q+p Which yields:

\displaystyle  \frac{1}{pq}     =  \frac{  -1}{x(p  + q + x)}

cross multiplication:

\displaystyle    x(p  + q + x)  =   - pq

distribute:

\displaystyle    xp  + xq +  {x}^{2} =   - pq

isolate -pq to the left hand side and change its sign:

\displaystyle    xp  + xq +  {x}^{2} + pq =  0

rearrange it to standard form:

\displaystyle   {x}^{2} +    xp  + xq  + pq =  0

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so

factor out x:

\displaystyle  x( {x}^{} +   p ) + xq  + pq =  0

factor out q:

\displaystyle  x( {x}^{} +   p ) +q (x + p)=  0

group:

\displaystyle  ( {x}^{} +   p ) (x + q)=  0

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

\displaystyle   \begin{cases} \displaystyle  {x}^{} +   p  = 0 \\   \displaystyle x + q=  0 \end{cases}

cancel out p from the first equation and q from the second equation which yields:

\displaystyle   \begin{cases} \displaystyle  {x}^{}   =  - p \\   \displaystyle x  =  -  q \end{cases}

and we are done!

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2 years ago
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PolarNik [594]

Answer:

Explanation is in a file

Step-by-step explanation:

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2 years ago
Question 7 of 10
Mama L [17]

Answer:

B

Step-by-step explanation:

Given a quadratic equation in standard form

ax² + bx + c = 0 ( a ≠ 0 )

Then the discriminant is

Δ = b² - 4ac

• If b² - 4ac > 0 then 2 real irrational roots

• If b² - 4ac > 0 and a perfect square then 2 real rational roots

• If b² - 4ac = 0 then 1 real double root

• If b² - 4ac < 0 then 2 complex roots

Given

x² + 3x - 7 = 0 ← in standard form

with a = 1, b = 3, c = - 7 , then

b² - 4ac

= 3² - (4 × 1 × - 7) = 9 + 28 = 37

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Georgia [21]

-7x + 12 is not an equiv. expression because one of the - signs in the original expression has been changed to +.

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