# Limit at infinity problem

1. Aug 16, 2007

### Ionophore

Can someone give me a hint on how to evaluate the following limit?

$$\stackrel{lim}{T\rightarrow\infty} (Texp(c/T) - T)$$

I tried multiplying the numerator and denominator by the conjugate (because that sometimes helps) and got:

$$(T^2exp(2c/T) - T^2) / (Texp(c/T) + T)$$

But I'm not sure what I can do from there...

2. Aug 16, 2007

### nicktacik

You can express it as

$$\lim_{x\rightarrow\infty} \frac{e^{c/x} - 1}{x^{-1}}}$$

Then apply l'hopital's rule

3. Aug 16, 2007

### Curious3141

Even faster, just observe that c/T -> 0 as T-> inf and use Maclaurin's for e^(c/T) up to the first order term.

4. Aug 16, 2007

### Hurkyl

Staff Emeritus
Here's a typesetting tip:

\lim_{x \to a}

results in

$$\lim_{x \to a}$$

Furthermore, if you wanted to create your own custom one, you would do this:

\mathop{\mathrm{Hur}}_{a = 1}^{b = 7}

to get

$$\mathop{\mathrm{Hur}}_{a = 1}^{b = 7}$$