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#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

- Thread starter karush
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- #1

- Jan 31, 2012

- 2,678

Why are you plugging in large negative values for x? Surely for an infinite limit you should be plugging in large positive values.Ok all I did was DesmosNot real sure how to take limit

As for a hint, you should use the standard limit $\displaystyle \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right) ^x = \mathbf{e} $.

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- Jan 31, 2012

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Ok I see what you mean

But there is no graph on the positive side

also its 2 not 1

But there is no graph on the positive side

also its 2 not 1

Last edited:

- Jan 29, 2012

- 1,151

What do you mean "there is no graph on the positive side"? Of couse there is.

- Aug 30, 2012

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Prove It is not giving you the answer he is giving you a suggestion that you can use the limit he posted. See if there is any kind of substitution you can make to put your limit into the form he gave you.Ok I see what you mean

But there is no graph on the positive side

also its 2 not 1

And the graph of f(x) becomes real again for \(\displaystyle x \geq 2\). (Why does it "disappear?" Why does it "reappear?")

-Dan

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- Jan 31, 2012

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actually I don't know why it does not graph $0\le x \le 2$

Look at the numbers. Can you divide by 0? Can you take an even root of a negative number?actually I don't know why it does not graph $0\le x \le 2$

- Jan 29, 2012

- 1,151

Using the "Desmos graphing calculator" at [FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif]https://www.desmos.com/calculator you can look at the graph at very large x and small values of y so get an idea of the values you need.[/FONT]