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3 years ago

Ayuda en matemáticas plis

Mathematics

1 answer:

Semenov [28]3 years ago

3 0

Answer:

A-2515.57647469

Step-by-step explanation:

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Calculate the limit values:

Nataliya [291]

A) This particular limit is of the indeterminate form,
$\frac{ \infty }{ \infty }$
if we plug in infinity directly, though it is not a number just to check.

If a limit is in this form, we apply L'Hopital's Rule.

's
$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }$
So we take the derivatives and obtain,

$Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}$

Still it is of the same indeterminate form, so we apply the rule again,

$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x}$

This simplifies to,

$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0$

b) This limit is also of the indeterminate form,

$\frac{0}{0}$
we still apply the L'Hopital's Rule,

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }$

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 }$

When we plug in zero now we obtain,

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1$
c) This also in the same indeterminate form

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }$

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }$

It is still of that indeterminate form so we apply the rule again, to obtain;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }$

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }$

This gives us;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2$

d) $Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x$

For this kind of question we need to rationalize the radical function, to obtain;

$Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}$

We now divide both the numerator and denominator by x, to obtain,

$Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}$

This simplifies to,

$=\frac{2}{\sqrt{1+0}+1}=1$

5 0

3 years ago

How to write y= 2x-7 in function notation

luda_lava [24]

I believe it would be <span>f(x)=2x-7</span>

4 0

3 years ago

Read 2 more answers

2 What is the solution set of <br> |4x - 1|= 3?

myrzilka [38]

Answer:

Exact Form:

x = 1 , − 1/ 2

Decimal Form:

x = 1 , − 0.5

Step-by-step explanation:

QUICK MATH

6 0

3 years ago

The following numbers of people attended the last 10 screenings of a movie.

sdas [7]

110 and 294 are the outliers because their difference in value from other data is large.

<h3>What are outliers?</h3>

An outlier is an observation that lies an abnormal distance from other values in a random sample from a population.

<h3>sample of data</h3>

Given the sample data 110, 196, 197, 199, 205, 208, 209, 210, 210, 294, we need to get the outliers that are an observation that lies an abnormal distance.

From the data, we can see clearly that 110 and 294 are the outliers. Their difference in value from other data is large.

Learn more on outliers here: brainly.com/question/2749543

6 0

3 years ago

Kathy buys a bicycle after a 30% markdown. The original price was $490.00. What did<br> she pay?

Umnica [9.8K]

Answer:

$343

Step-by-step explanation:

To get 30% of 490 we divide 30% by 1

30% ÷ 1 = 147

Now, we subtract 490 by 147

490 - 147 = 343

4 0

3 years ago