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luda_lava [24]
2 years ago
9

Please help it’s a timed exam

Mathematics
1 answer:
lina2011 [118]2 years ago
7 0

Answer:

Here 86 is answer.

So you better choose 86.

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Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0
Vsevolod [243]

Answer:

\dfrac{-1}{6}

Step-by-step explanation:

Given the limit of a function expressed as \lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}, to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)

Step 2: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the function

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\

Step 3: substitute x = 0 into the resulting function

= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)

Step 4: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\

=  \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)

Step 6: Apply  L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\

Step 7: substitute x = 0 into the resulting function in step 6

=  \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}

<em>Hence the limit of the function </em>\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \  is \ \dfrac{-1}{6}.

3 0
3 years ago
Evaluate P(3, 3) <br> 1<br> 9<br> 6
solong [7]
Hence, the value of P(3,3) is 6
5 0
2 years ago
Can someone help me please
goldenfox [79]

Answer:

1. \quad\dfrac{1}{k^{\frac{2}{3}}}\\\\2. \quad\sqrt[7]{x^5}\\\\3. \quad\dfrac{1}{\sqrt[5]{y^2}}

Step-by-step explanation:

The applicable rule is ...

  x^{\frac{m}{n}}=\sqrt[n]{x^m}

It works both ways, going from radicals to frational exponents and vice versa.

The particular power or root involved can be in either the numerator or the denominator. The transformation applies to the portion of the expression that is the power or root.

7 0
3 years ago
What is the measure of &lt;2?
Ugo [173]
I think 90 degrees but not 100 percent sure
5 0
3 years ago
............°_°.............​
Rasek [7]

Answer:

3/4 and 0.95 would both result in a reduction.

Step-by-step explanation:

multiplying by anything less than 1 will make it smaller which both 3/4 and 0.95 are smaller than 1.

8 0
2 years ago
Read 2 more answers
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