Evaluate the common limit

Question

I'd like more of an explanation than a solution, I'm sorry but I'm studying math again after 12 years, and I don't understand basic concepts.

I have to evaluate some limits without using L'Hopital rule, just by definition. So I got:

$$\lim\limits_{h\to \:0}\left(\frac{e^h-1}{h}\right).$$

After evaluating it in a Math app, it states that it's just 1 due to the common limit evaluation, but I don't understand why it's 1.

What's your definition of $e^h$? In any case, your limit is well-scrutinized here. — Sarvesh Ravichandran Iyer, May 18 '22 at 09:19
One can develop a very good intuitive understanding of $e^h$ for small $h$ by noticing that $1+h$ is a fantastic approximation. — Kurt G., May 18 '22 at 12:26

score 1 · Accepted Answer · answered May 18 '22 at 05:06

1

Depends what your definition of $e^h$ is.

If you know the power series ($e^h =\sum_{n=0}^{\infty} \frac{h^n}{n!}$), it is easy.

If you know that $(e^h)' = e^h$, then $e^h-1 =\int_0^h e^x dx $, and you can use the mean value theorem combined with $e^0=1$ and the continuity of $e^h$ to get the result.

There are probably a few other ways, but these occurred to me.

answered May 18 '22 at 05:06

marty cohen

107,799

3

Even easier, by definition the derivative of $e^h$ at $h=0$ is the desired limit. – J.G. May 18 '22 at 05:16
1

Very true. It's nice when problems have lots of different solutions. – marty cohen May 18 '22 at 14:43

Evaluate the common limit

1 Answers1