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IgorC [24]
2 years ago
6

Evaluate the limit

Mathematics
2 answers:
Digiron [165]2 years ago
6 0

We can transform the limand into a proper rational expression by substitution.

Let y = √x. Then as x approaches 4, y will approach √4 = 2. So

\displaystyle \lim_{x\to4}\frac{\sqrt x - \sqrt{3 \sqrt x - 2}}{x^2 - 16} = \lim_{y\to2} \frac{y - \sqrt{3y-2}}{y^4 - 16}

Now let z = √(3y - 2). Then as y approaches 2, z will approach √(3•2 - 2) = 2 as well. It follows that y = (z² + 2)/3, so that

\displaystyle \lim_{y\to2} \frac{y - \sqrt{3y-2}}{y^4-16} = \lim_{z\to2} \frac{\frac{z^2+2}3 - z}{\frac{(z^2+2)^4}{81}-16} \\\\ = \lim_{z\to2} \frac{27(z^2+2)-81z}{(z^2+2)^4 - 1296} \\\\ = 27 \lim_{z\to2} \frac{z^2 - 3z + 2}{z^8 + 8z^6 + 24z^4 + 32z^2 - 1280}

Plugging z = 2 into the denominator returns a value of 0, which means z - 2 divides z⁸ + 8z⁶ + 24z⁴ + 32z² - 1280 exactly. Polynomial division shows that

\dfrac{z^8 + 8z^6 + 24z^4 + 32z^2 - 1280}{z-2} \\\\ = z^7+2z^6+12z^5+24z^4+72z^3+144z^2+320z+640

and it's easy to see that the numerator is also divisible by z - 2, since

z^2 - 3z + 2 = (z - 1) (z - 2)

So, we can eliminate the factor of z - 2 and we're left with

\displaystyle 27 \lim_{z\to2} \frac{z^2 - 3z + 2}{z^8 + 8z^6 + 24z^4 + 32z^2 - 1280} = 27 \lim_{z\to2}\frac{z-1}{z^7+\cdots+640}

The remaining limand is continuous at z = 2, so we can evaluate the limit by direct substitution:

\displaystyle 27 \lim_{z\to2}\frac{z-1}{z^7+\cdots+640} = \frac{27}{3456} = \boxed{\frac1{128}}

wel2 years ago
3 0

We are given with a limit and we need to find it's value so let's start !!!!

{\quad \qquad \blacktriangleright \blacktriangleright \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

But , before starting , let's recall an identity which is the <em>main key</em> to answer this question

  • {\boxed{\bf{a^{2}-b^{2}=(a+b)(a-b)}}}

Consider The limit ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

Now as directly putting the limit will lead to <em>indeterminate form 0/0.</em> So , <em>Rationalizing</em> the <em>numerator</em> i.e multiplying both numerator and denominator by the <em>conjugate of numerator </em>

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}\times \dfrac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-\sqrt{3\sqrt{x}-2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(x^{2}-4^{2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Using the above algebraic identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x})^{2}-(\sqrt{3\sqrt{x}-2})^{2}}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-(3\sqrt{x}-2)}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}+2}{\{(\sqrt{x})^{2}-2^{2}\}(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , here we <em>need</em> to <em>eliminate (√x-2)</em> from the denominator somehow , or the limit will again be <em>indeterminate </em>,so if you think <em>carefully</em> as <em>I thought</em> after <em>seeing the question</em> i.e what if we <em>add 4 and subtract 4</em> in <em>numerator</em> ? So let's try !

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2+4-4}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(x-4)+2+4-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , using the same above identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+6-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+3(2-\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take minus sign common in <em>numerator</em> from 2nd term , so that we can <em>take (√x-2) common</em> from both terms

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-3(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take<em> (√x-2) common</em> in numerator ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)\{(\sqrt{x}+2)-3\}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Cancelling the <em>radical</em> that makes our <em>limit again and again</em> <em>indeterminate</em> ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\cancel{(\sqrt{x}-2)}\{(\sqrt{x}+2)-3\}}{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}+2-3)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-1)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , <em>putting the limit ;</em>

{:\implies \quad \sf \dfrac{\sqrt{4}-1}{(\sqrt{4}+2)(4+4)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}}

{:\implies \quad \sf \dfrac{2-1}{(2+2)(4+4)(2+\sqrt{3\times 2-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{6-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{4})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+2)}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(4)}}

{:\implies \quad \sf \dfrac{1}{128}}

{:\implies \quad \bf \therefore \underline{\underline{\displaystyle \bf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\dfrac{1}{128}}}}

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40.51x+12.45y=666.64
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Answer:

The graph of the equation 40.51x+12.45y=666.64 is attached with the answer where the horizontal axis represents the X axis and the vertical axis represents Y axis.

To plot the graph physically just find two points lying on the line. Mark the points on the graph sheet and then join them. This will give you the line represented by the equation.

To find points on the line assume the value of any one variable, substitute it in the equation, then solve the equation to find the value of other variable. For example : assume y = 1; substitute the value of y in the equation;

⇒ 40.51x + 12.45×1 = 666.64

⇒ 40.51x = 666.64 - 12.45

⇒ 40.51x = 654.19

⇒ x = \frac{654.19}{40.51}

⇒ x ≈ 16.149

Therefore point ( 16.149 , 1 ) lie on the graph of the equation.

***Only two points are required to plot this graph just because it represents a straight line, that we can conclude just by observing the equation. If in an equation the power of x is 1 or 0 and power of y is 1 or 0 then only it will represent a straight line in 2-D plane.***

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In a triangle ABC, measure of angle B is 90 degrees. AB is 3x-2 units and BC is x+3. If the area of the triangle is 17 sq cm, fo
Scorpion4ik [409]

Answer:

x=\frac{8}{3}\ cm

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

A=\frac{1}{2}(AB)(BC)

we have

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AB=(3x-2)\ cm

BC=(x+3)\ cm

substitute the values

17=\frac{1}{2}(3x-2)(x+3)

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34=3x^2+9x-2x-6

3x^2+7x-6-34=0

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The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

3x^2+7x-40=0

so

a=3\\b=7\\c=-40

substitute in the formula

x=\frac{-7(+/-)\sqrt{7^{2}-4(3)(-40)}} {2(3)}

x=\frac{-7(+/-)\sqrt{529}} {6}

x=\frac{-7(+/-)23} {6}

x=\frac{-7(+)23} {6}=\frac{16}{6}=\frac{8}{3}

x=\frac{-7(-)23} {6}=-5

therefore

The solution is

x=\frac{8}{3}\ cm

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