limits

[Janet Ward]

f(x) = 9/2+e^(1/x)
I need to find the limit as X approaches 0-
I made a table and it appears that the limit is 3 but I keep getting awrong answer.
What am I doing wrong?

 

[soroban]

Hello, Janet Ward!

You are misreading that exponent.

\(\displaystyle \displaystyle \lim_{x\to0}\frac{9}{2 + e^{\frac{1}{x}}}\)

. . \(\displaystyle \begin{array}{c|c} x & f(x) \\ \hline \\[-2mm] 2 & \dfrac{9}{2+e^{\frac{1}{2}}} \\ \\[-2mm] 1 & \dfrac{9}{2+e} \\ \\[-2mm] \frac{1}{2} & \dfrac{9}{2 + e^2} \\ \\[-2mm] \frac{1}{3} & \dfrac{9}{2+e^3} \\ \vdots & \vdots \\ \\[-2mm] \frac{1}{100} & \dfrac{9}{2+e^{100}} \\ \vdots & \vdots \end{array}\)

 

[lookagain]

f(x) = 9/(2+e^(1/x))

I need to find the limit as X approaches 0- \(\displaystyle \text{This is the limit as x approaches 0 from the left-hand side.}\)
I made a table and it appears that the limit is 3 but I keep getting awrong answer.
What am I doing wrong?

Hello, Janet Ward!

You are misreading that exponent.

\(\displaystyle \displaystyle \lim_{x \to 0^{-}} \frac{9}{2 + e^{\frac{1}{x}}}\)

. . \(\displaystyle \begin{array}{c|c} x & f(x) \\ \hline \\[-2mm] -2 & \dfrac{9}{2+e^{-\frac{1}{2}}} \\ \\[-2mm] -1 & \dfrac{9}{2+e^{-1}} \\ \\[-2mm] \frac{-1}{2} & \dfrac{9}{2 + e^{-2}} \\ \\[-2mm] \frac{-1}{3} & \dfrac{9}{2+e^{-3}} \\ \vdots & \vdots \\ \\[-2mm] \frac{-1}{100} & \dfrac{9}{2+e^{-100}} \\ \vdots & \vdots \end{array}\)

\(\displaystyle \frac{9}{2 + 0} =\)

\(\displaystyle \frac{9}{2}\)

 

[Janet Ward]

Thank you

 

[Janet Ward]

I'm having trouble estimating the limit

lim (1+x)^1/x
as x approaches 0
I made a table and can' determine the limit.
Am I using incorrect x values?

 

[galactus]

This is a 'famous' limit.

Use smaller and smaller values in your chart.

Say, 1.

Then, .1

Then, .01

and so forth, getting closer to 0.

Also, try it from the other direction.

You will see it converges to a rather famous constant that pops up everywhere in the mathworld.

\(\displaystyle \begin{array}{c|c}x&\text{value}\\ \hline 1&2\\ 0.1& 2.5937424601\\ .01&2.70481382942\\ .001&2.71692393224\end{array}\)

Can you see what it is converging to?.

 

[Janet Ward]

It appears to be approching 2.72 or do I round up to 3?

 

[JeffM]

It appears to be approching 2.72 or do I round up to 3?

Galactus gave two BIG hints in his last post.

Did you try what he suggested about trying the other direction?

x value

-.1 2.868
-.01 2.732
-.001 2.7296

You certainly cannot round up to 3. Do you see why?

Now consider his other hint.

 

[Janet Ward]

I skipped over his suggestion to try the other limit. i narrowed it down to 2.71828. Thanks for your help. I'll try to pay more attention next time!

 

[galactus]

Your valuE of 2.71... is thE numErical valuE of a vEry famous constant. Hint, Hint, Hint.....

 

[Deleted member 4993]

You mEan Euler's constant???

 

[lookagain]

You mEan Euler's constant???

That is not Euler's constant; it is "Euler's number."

Euler's constant is the limit (as n --> oo) of [(1 + 1/2 + 1/3 + ... + 1/n) - ln(n)].

It is approximately 0.5772.


Euler's number is e, which is about 2.718.

For the difference, please see:

http://en.wikipedia.org/wiki/Euler%E2%8 ... i_constant

 

[JeffM]

Lookagain rEally mEant to say EulEr's numbEr.

 

[lookagain]

Lookagain rEally mEant to say EulEr's numbEr.

I did not mean anything else.

Please go back and read my more recent post, as well about
the first dozen lines in the link I gave.

 

[Deleted member 4993]

You mEan Euler's constant???

That is not Euler's constant; it is "Euler's number."

I stand corrected.

I am really glad that we have a vigilant editor among us.