Help needed calculating the limit of a function. Thanks in advance

**Real9999** · September 29th 2010, 01:09 PM

I am struggling in calculating the limit of the following function

When x approaching zero from the right (0+),

lim (lnx/e^2x)

The problem I have here is how to handle lnx because lnx is not defined at 0

Please help and thanks!

**HappyJoe** · September 29th 2010, 01:24 PM

First of all, recall that ln(x) tends towards -infinity for x->0+.

For x<1, we have e^(2x) < e^2, hence 1/e^(2x) > 1/e^2.

For x<1, ln(x) is negative, so multiplying both sides of the inequality by ln(x) gives ln(x)/e^(2x) < ln(x)/e^2.

And then you use my initial reminder, which says that the right hand side of this last inequality will tend to -infinity, and hence so will the left hand side.

**Real9999** · September 30th 2010, 05:14 AM

thanks!!!!!

**Archie Meade** · September 30th 2010, 05:36 AM

Originally Posted by Real9999

I am struggling in calculating the limit of the following function

When x approaching zero from the right (0+),

lim (lnx/e^2x)

The problem I have here is how to handle lnx because lnx is not defined at 0

Please help and thanks!

The limit of a quotient is the quotient of the limits,
provided that the limit of the denominator is not zero.

$\displaystyle\lim_{x \to 0+}\frac{log_ex}{e^{2x}}=k\Rightarrow\frac{\displa ystyle\lim_{x \to 0+}log_ex}{\displaystyle\lim_{x \to 0+}e^{2x}}=k$

$e^0=1$

$\displaystyle\lim_{x \to 0+}log_ex=k\Rightarrow\lim_{x \to 0+}x=e^k$

hence $k\rightarrow\ -\infty$

so there is no limit.

**HallsofIvy** · September 30th 2010, 06:01 AM

That's not a bad "heuristic" way of doing it but I suspect many teachers would reject it as a valid argument. "The limit of a quotient is the quotient of the limits, provided that the limit of the denominator is not zero" and both limits exist. In this case, the limit of the numerator does not exist. Saying that a limit "is infinity" or "is negative infinity" is just saying the limits do not exist for a particular reason. Happy Joe's argument is better.

**Real9999** · September 30th 2010, 06:48 AM

Yup, I agree that Happy's response is much better.

The trick here is to transform the original function to one that is more tractable rather than computing the limit directly...

Thanks!!!!!!!!

**Archie Meade** · September 30th 2010, 07:44 AM

Originally Posted by Archie Meade

The limit of a quotient is the quotient of the limits,
provided that the limit of the denominator is not zero.

$\displaystyle\lim_{x \to 0+}\frac{log_ex}{e^{2x}}=k\Rightarrow\frac{\displa ystyle\lim_{x \to 0+}log_ex}{\displaystyle\lim_{x \to 0+}e^{2x}}=k$

$e^0=1$

$\displaystyle\lim_{x \to 0+}log_ex=k\Rightarrow\lim_{x \to 0+}x=e^k$

hence $k\rightarrow\ -\infty$

so there is no limit.

$k\rightarrow\ -\infty$ means $k$ approaches $-\infty$

which in turn means that $k$ is not a constant,

hence the quotient has no limit.

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