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1 year ago

If the limit of f(x) as x approaches 8 is 3, can you conclude anything about f(8)?

1 answer:

6 0

If the limit of f(x) as x approaches 8 is 3, can you conclude anything about f(8)? The answer is No. We cannot. See the explanation below.

<h3>What is the justification for the above position?</h3>

Again, 'No,' is the response to this question. The justification for this is that the value of a function does not depend on the function's limit at a given moment.

This is particularly clear when we consider a question with a gap. A rational function with a hole is an excellent example that will help you answer this question.

The limit of a function at a position where there is a hole in the function will exist, but the value of the function will not.

<h3>What is limit in Math?</h3>

A limit is the result that a function (or sequence) approaches when the input (or index) near some value in mathematics.

Limits are used to set continuity, derivatives, and integrals in calculus and mathematical analysis.

Learn more about limits:
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If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o

almond37 [142]

If the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve $=\int\limits^a_b {x} \, dx$

We have,

$y^2 =4ax$

⇒ $y=\sqrt{4ax}$

$x= 4a$ ,

Area of the region $=\frac{256}{3}$ Sq units

Now comparing both given equation to get the intersection between points;

$y^2=16a^2$

$y=4a$

So,

Area bounded by the curve $= \[ \int_{0}^{4a} y \,dx \]$

$\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]$

$\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]$

$\frac{256}{3}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}$

On applying the limits we get;

$\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} * 8 *a^{2} \sqrt{a}$

$\frac{256}{3}= \frac{4}{3} * 8 *a^{3}$

⇒ $a^{3} =8$

$a=2$

Hence, we can say that if the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

To know more about Area bounded by the curve click here

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8 0

1 year ago

Read 2 more answers

SOLVE X FOR THE EQUATION X/2=17 (PLEASE HELP ME)

NeX [460]

To get x to be by itself you would have to mutiply both sides by two so then it would look like this x=34. So the answer to your question is x=34

3 0

3 years ago

I need help answering number 8 please

swat32

The diagram of the triangle is shown below

One interior angle is 56°
One exterior angle is 103° ⇒ we can work out one interior angle 180°-103° = 77°

The other interior angle is 180° - (56°+77°) = 44°

5 0

3 years ago

NO LINKS!! Please help me with these notes. Part 1a

erik [133]

Answers:

When we evaluate a logarithm, we are finding the exponent, or <u> power </u> x, that the <u> base </u> b, needs to be raised so that it equals the <u> argument </u> m. The power is also known as the exponent.

$5^2 = 25 \to \log_5(25) = 2$