N + 5 ≥ 25

Mathematics

1 answer:

Triss [41]3 years ago

4 0

The answer is A. Subtract the five on both sides and you end up with A :)

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

e-lub [12.9K]

$sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0$

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 log(x)=ln(x).

The limit result can be proven if the base of log(x) is 10.

$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

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

 $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$

Therefore,

 $\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}$

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3 years ago

Jet001 [13]

Answer:

It is A

Step-by-step explanation:

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4 years ago

Mary wants to make cookies. To make cookies she needs half a cup of flour per batch of cookies. If Mari has 5 cups of flour how

Anit [1.1K]

10!

There's two halves in a whole. So we could do 5 x 2 to get our answer. Hope this helps!

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3 years ago

Simplify the expression (4i)(2i)(-9i)

lakkis [162]

Answer:

Step-by-step explanation:

(4i)*(2i)*(-9i) = 4*2*(-9) * i³ { i³ = -i}

= -72 *(-i)

= 72i

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What is the equation of the following line? Be sure to scroll down first to see all answer options.

ivolga24 [154]

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The equation is...
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3 years ago

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