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Lelechka [254]
3 years ago
10

Find the limit of the function algebraically. (2 points) limit as x approaches zero of quantity x squared plus two x divided by

x to the fourth power.

Mathematics
1 answer:
DIA [1.3K]3 years ago
7 0

Answer:

DNE

Step-by-step explanation:

\lim_{x \to \0} \frac{x^{2} +2x}{x^{4} }

As you can see in the picture I attached to this, that as the limit goes to zero from the negative side, it approaches -∞ and from the positive side, it approaches ∞ . hence, the limit doesn't exist.

To show this algebraically, you have to imagine how a number divided by zero looks like, Just know the graph of 1/x and see how the limit to zero doesnot exist. I hope this helps!

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Step-by-step explanation:

With this kind of problem, we're looking at an equation in the form

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Help me with this please!!
valina [46]
<h2><u>Given:</u><u>-</u></h2>

  • Points C = (-7,2) → \sf{(X_1,Y_1)}
  • D = (3,12) → \sf{(X_2,Y_2)}

<h2><u>To </u><u>Find</u><u>:</u><u>-</u></h2>

  • The Midpoint of CD.

<h2><u>Required</u><u> </u><u>Response</u><u>:</u><u>-</u></h2>

Let,

Midpoint of CD be (x,y).

WKT,

\boxed{\sf{(x,y) = \frac{X_1+X_2}{2},\frac{Y_1+Y_2}{2}}}

→\;{\sf{\frac{-7+3}{2},\frac{2+12}{2}}}

→\;{\sf{\frac{-4}{2},\frac{14}{2}}}

→\;{\sf{-2,7}}

The Midpoint of CD ◕➜ \Large{\red{\mathfrak{(-2,7)}}}

Let,

The centre be O

Radius = CO & OD

Here, C = (-7,2) → \sf{(X_1,Y_1)}

O = (-2,7) → \sf{(X_2,Y_2)}

\boxed{\sf{Distance = \sqrt{(X_2-X_1)²+(Y_2-Y_1)²}}}

→\;{\sf{\sqrt{(-2+7)²+(7-2)²}}}

→\;{\sf{\sqrt{5²+5²}}}

→\;{\sf{\sqrt{25+25}}}

→\;{\sf{\sqrt{50}}}

→\;{\sf{5√2 (or) 7.07}}

Radius of Circle ◕➜ \Large{\red{\mathfrak{7.07}}}

<h2>Option D.</h2>

Hope It Helps You ✌️

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