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1. Hi all, I have an Engineering interview at Oxford this Monday and I am currently trying to practice for it. I see they often give a funtion and ask you to graph for it. Im just looking for general advice as to how to go about this.

Things I know to do...

- find where it cuts x and y axis
- look for turning points
- limits??

In what order should I first attempt these or does it not matter?

Here is a more specific example

Take y= (sinx) / x (no idea how to write in proper maths form)

I don't know how to go about arriving at this graph...

Can someone please talk me through it if you can? I don't understand the oscillations decreasing as x increases/decreases towards infinity, and

What are limits and how can I go about calculating them? I havent covered this yet.

Thanks all, I hope someone can help with at least something
2. Key things to look for are:

- where it intercept the axes (if possible)
- turning points
- asymptotes/limits

- behaviour as x (and y) tends to positive and negative infinity
- other special features

In the sinx/x example, do you know the small angle approximation? It states that for small x, sinx is roughly equal to x, so when x = 0 sinx/x is roughly 1.

The points where it intercepts the x-axis can be found by noting what values for x make sinx = 0
Spoiler:
Show
Observe that sinx is a periodic function with period

Turning points can be found by deriving the function, but can't be found very easily numerically. Try plugging in a few values.

Special features: Note that sinx alternates between 1 and -1. so if we consider the graphs of 1/x and -1/x we have an envelope (we are always multiplying 1/x by a number so 1/x and -1/x 'envelope' the function, see Envelopes. Notice that the peaks appear to decrease in a similar way to 1/x and -1/x).

As x tends to positive and negative infinity, what do you notice?
3. You could see it as multiplying sin(x) by 1/x. sin(x) just oscillates between -1 and 1, and 1/x tends to zero as x goes to infinity. In terms of limits, the limit of 1/x as x tends to infinity can be written as:

lim x->infinity (1/x) = 0.

This is the same for negative values of x, o the oscillations decrease as you move away from x=0. You could also see it as when |x| increases, you're dividing by an ever increasing value, so the oscillations of f(x) become smaller and smaller.
4. Wow thanks guys, that's really a lot of help. + rep.
5. (Original post by Goldfishy)
When x = 0 sinx/x is roughly 1.
0/0 is not 1.
6. (Original post by j.alexanderh)
0/0 is not 1.
I never said it was. I said that:

...for small x, sinx is roughly equal to x, so when x = 0 sinx/x is roughly 1.
...implying that sinx/x tends to 1 as x tends to 0. I also explicitly mentioned the small angle approximation, to avoid the ambiguities of 0/0

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Updated: December 5, 2010
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