Exponential Fundamental Limits without using L'Hôspital's rule

Question

I have a limit to evaluate.

$$\lim_{x\to2} \left(\frac{\mathrm e^x - \mathrm e^2}{x-2}\right)$$

Can someone solve it without using L'Hôspital and explain me the steps?

Thanks

Answer 1

Do the substitution $x=t+2$, so the limit becomes $$ \lim_{t\to0}\frac{e^{t+2}-e^2}{t}=e^2\lim_{t\to0}\frac{e^t-1}{t} $$ The final limit is standard: if you're not allowed to use l'Hôpital, this limit should be allowed to use.

Answer 2

Recall the definition of "derivative": $f'(a) = \lim\limits_{x\to a} \dfrac{f(x)-f(a)}{x-a}$.

Now let $f(x) = e^x$ and $a=2$. Then you have $f'(x)=e^x$ so $f'(2) = e^2$ and so $$ e^2 = f'(2) = \lim_{x\to 2} \frac{e^x-e^2}{x-2}. $$

Answer 3

One could use the mean value theorem as well:

$$ \frac{e^{x}-e^2}{x-2} = \frac{e^{\xi(x)}(x-2)}{x-2}=e^{\xi(x)}$$

for some $\xi$ between $x$ and 2. Hence, $\lim_{x\rightarrow 2} \xi(x)=2$ and thus

$$ \lim_{x\rightarrow 2} \frac{e^{x}-e^2}{x-2} = \lim_{x\rightarrow 2} e^{\xi(x)} = e^2.$$