How to evaluate the limit $\lim_{h \to 0} \frac{e^{2+h}-e^2}h$?

Question

$$ {\lim \limits_{h \to 0}} { {e^{2+h}-e^2 } \over {h} } $$

Due to time constraints, evaluating limits with e in them wasn't covered and I have this on the AP exam review. How do I proceed?

Answer 1

At the beginning of the year, in your AP Calculus AB class, you found derivatives by using one of two definitions:

1) The derivative of a function at a specific x-value: $\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a)$

2) The derivative of a function "in general", i.e. the "derivative function": $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=f'(x)$

The second formula gives the derivative at a general $x$-value. But, if you plug in a value for $x$ in the formula (such as $2$), then you can get a formula, similar to #1, that gives the derivative of a function at a specific $x$-value:

3) $\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}=f'(a)$

So, for your problem, notice that it is the form of the formula #3...your $f(x)=e^x$, and your $a$ value is $2$. Then, formula #3 becomes:

$\lim_{h\rightarrow 0}\frac{f(2+h)-f(2)}{h}=f'(2)$

$\lim_{h\rightarrow 0}\frac{e^{2+h}-e^2}{h}=f'(2)$

Note that, for $f(x)=e^x$, $f'(x)=e^x$, so:

$\lim_{h\rightarrow 0}\frac{e^{2+h}-e^2}{h}=e^2$

Answer 2

Hint. One may recall that for a differentiable function around $x_o$ we have $$ \frac{f(x_0+h)-f(x_0) }{h} \to f'(x_0),\quad h\to 0. $$ Apply it to $f(x)=e^x$.

Answer 3

Hint: If $f(x)=e^x$, then the given limit is $f'(2)$.

Answer 4

$$\lim_{h\to 0}\frac{e^{2+h}-e^2}{h}=e^2\cdot\lim_{h\to 0}\frac{e^h-1}{h}=e^2\cdot 1= e^2$$

This follows from the fact that,

$$e^x=1+x+\mathcal{O}(x^2)\implies e^x-1=x+\mathcal{O}(x^2)\implies \frac{e^x-1}{x}=1+\mathcal{O}(x)$$

Answer 5

Let's see what it boils down to. You can rewrite it as:

$$\lim_{h\rightarrow 0}\frac{e^{2+h}-e^2}{h}=\lim_{h\rightarrow 0}\frac{e^{2}e^{h}-e^2}{h}=e^2\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}.$$

So you only need to figure out this last limit. If you are allowed to make use of knowledge about the derivative of $e^x$, then you are done, because the last limit is this derivative evaluated at $0$.

If you are not allowed to do that, I suggest using a Taylor series expansion of $e^x$. Have you covered that yet?

Answer 6

$${\lim \limits_{h \to 0}} { {e^{2+h}-e^2 } \over {h} } = \frac{d}{dx}\left(e^x\right)\Big|_{x=2}.$$

the left hand side is the limit definition of the derivative of $e^x$ at $x = 2$, then right hand side. use $(e^x)' = e^x.$