- anonymous

I am trying to sketch the curve y=x+1/x
Following are the steps that are generally followed. But I don’t understand one thing
First I note that while as x tends to infinity, 1/x tends to 0. Then we conclude that y tends x, since 1/x tends to 0. The thing that I don’t understand is if x tends to infinity, so why don't we say y tends to infinity too.

- katieb

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- amistre64

y = (x+1)/x ? or y = x + (1/x)

- anonymous

No as it is

- anonymous

I mean using the math rule

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## More answers

- anonymous

First division then addition...................

- amistre64

id say its y = inf + 0 as well...... is there an answer that says differently?

- anonymous

Its curve sketching. If I say that way, the curve turns out to be wrong

- anonymous

iamignorant,
Is the following your expression?
\[y=x+\frac{1}{x} \]

- anonymous

Yes

- anonymous

\[y==1+\frac{1}{x} \]
The 2 attached plots should help visualize the function.
The first plot is from x=-10 to +10. The scale of the plot is too large and should be reduced in your pdf viewer.
The second is from x= -4000 to x= +4000. To the the casual observer the plot looks like a straight line drawn through the origin and making an angle of Pi/4 with the x axis.

##### 2 Attachments

- anonymous

Just reviewed what was put on the net. The plots seem to be OK.

- anonymous

Thanks robtobey

- anonymous

You see the problem that is disturbing me is a little different

- anonymous

I am trying to sketch the curve using the asymptote location method

- anonymous

And I have presented the problem that I face while reasoning

- anonymous

I understand that while 1/x tends to infinity y must tend to x, or else y=x will not be an asymptote

- anonymous

But I don't understand where exactly I mis-reasoned to make it otherwise

- amistre64

y = (x^2 +1)/x is an equivalent equation right?

- anonymous

Yes sure

- amistre64

then the asymptote is a slant/oblique right?

- anonymous

Sorry, but how did you conclude that?

- amistre64

the ends approach y=x

- anonymous

Hold on a few minutes. I want to set up some Limits.

- amistre64

the top is one degree highr than the bottom

- anonymous

Actually I am not aware of that rule

- amistre64

when we divide it all by x we get:
x +1/x
------
1

- anonymous

Yes

- amistre64

anyting with an x on the bottom goes to zero; which leaves us with y = x as the slant asymptote

- anonymous

You mean when 1/x tends to zero anyting with an x on the bottom goes to zero

- anonymous

Is that what you mean?

- amistre64

so you have 2 curves that trail in the ends to the line y=x; and in the middle skirt along the x=0 axis: like this

##### 1 Attachment

- anonymous

Good job amistre64. I have to break off. Thank you.

- amistre64

when the bottom of a fraction gets large; the value gets small:
1
-------------- = .000000...00001
1000000...0000

- amistre64

so anything with an x on the bottom; as the x gets large; the value goes to zero.

- amistre64

whats left over is what your asymptote is :)

- anonymous

Actually, when x is tending towards infinity 1/x tends to zero (this is what you say), but x is tending to infinity so y=x+0 means y is tending to infinity.
This is what I don't understand

- amistre64

y is keeps going up and skirting next to the y=x line; instead of a y=3 line, its the y=x line. that line has a range of (-inf,inf)

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