Struggling with a Limit: Help Appreciated!

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SUMMARY

The limit problem presented is lim(X->0) 1/(ix)*(exp(imx) - 1) = m. The correct approach involves applying L'Hôpital's rule, which is valid since both the numerator and denominator approach zero as x approaches zero. By differentiating the numerator, e^{imx} - 1, and the denominator, ix, the limit simplifies to m as x approaches zero. The final result confirms that the limit evaluates to m.

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I can't get the following limit to work:

lim(X->0) 1/(ix)*(exp(imx) - 1 ) = m

I'm sorry for the poor notation. I tried expanding the exponential, and l'hospital's rule and combinations of these approaches, but i can't get it to work out. Any help is much appreciated!
 
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student111 said:
I can't get the following limit to work:

lim(X->0) 1/(ix)*(exp(imx) - 1 ) = m

I'm sorry for the poor notation. I tried expanding the exponential, and l'hospital's rule and combinations of these approaches, but i can't get it to work out. Any help is much appreciated!

I assume you mean [itex]\lim_{x\fo 0}\frac{e^{imx}- 1}{ix}[/itex]. I can see no reason why L'Hopital's rule would not work:

Both [itex]e^{imx}-1[/itex] and ix go to 0 as x goes to 0.

[itex]\left(e^{imx}\right)'= I am e^{imx}[/itex] while (ix)'= i. By l'hospital's rule, the limit is the same as [itex]\lim_{x\to 0}\frac{im e^{imx}}{i}= m\left(\lim_{x\to 0}e^{imx}\right)= m[/itex].
 
doh! I differentiated the whole part instead of nominator and denominator separately..

Thx a lot
 

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