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Anna71 [15]
3 years ago
8

Hello there!

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
6 0
Answers:
The limit as x approaches 3 does not exist (DNE)

The function value f(3) is equal to 5, so f(3) = 5

In short, the answer is choice B

=============================================

Explanations:

Let's start with computing the limit. First locate 3 on the x axis. Then move slightly to the left of 3, say to x = 2. Draw a vertical line upward until you hit the function curve. Mark the point on the function curve and then drag that point closer and closer to x = 3. Notice how y is getting loser to y = 3.

Then do the same for the other side of x = 3. Start at x = 4 and move to the left to get to x = 3. Get closer and closer, and you'll notice that y is getting closer to y = 5. These two differing y values tell us that the limit as x approaches 3 does not exist. 

Alternatively: The left hand limit (LHL) and right hand limit (RHL) are different, so the overall limit does not exist. 

---------------------

The function value f(3) is simply 5 because we draw a vertical line through x = 3 and it intersects the function at (3,5). Take note how I'm focusing on the closed circle and not the open circle. The open circle is a gap or hole in the graph.

Note: because the limit at x = 3 and the function value at x = 3 differ, this means we have a discontinuity. 
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5 0
2 years ago
Read 2 more answers
30+ Points!!!<br> 5. Solve the following inequalities.<br> a) 2 log3x – 2 logx3 -3 &lt;0
Ilya [14]

Answer:

I answered your last question also

2 log3x – 2 logx3 -3 <0

\mathrm{Subtract\:}2\log ^3\left(x\right)\mathrm{\:from\:both\:sides}

2\log ^3\left(x\right)-2logx^3-3-2\log ^3\left(x\right)

\mathrm{Simplify}

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\mathrm{Add\:}3\mathrm{\:to\:both\:sides}

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\mathrm{Simplify}

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Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)

\left(-2logx^3\right)\left(-1\right)>-2\log ^3\left(x\right)\left(-1\right)+3\left(-1\right)

\mathrm{Simplify}

2lx^3og>2\log ^3\left(x\right)-3

\mathrm{Divide\:both\:sides\:by\:}2lx^3o;\quad \:l>0

\frac{2lx^3og}{2lx^3o}>\frac{2\log ^3\left(x\right)}{2lx^3o}-\frac{3}{2lx^3o};\quad \:l>0\\

\mathrm{Simplify}

g>\frac{2\log ^3\left(x\right)-3}{2lx^3o};\quad \:l>0

Step-by-step explanation:

6 0
2 years ago
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