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

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?

Martin Sleziak
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Bassinator
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6 Answers6

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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 $f(x)=e^x$.

Olivier Oloa
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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:

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

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

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

This follows from the fact that,

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

Prasun Biswas
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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?

Mankind
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  • We have not covered taylor series, no. – Bassinator May 02 '15 at 18:57
  • Could you elaborate on the last step? – Bassinator May 02 '15 at 18:59
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    Yes, you somehow need to show that $\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}=1$. Perhaps you have a result that says that $\lim_{h\rightarrow 0} e^h/(1+h) = 1$, so you are allowed to substitute $e^h$ in the limit with $1+h$. This result stems from a Taylor expansion of $e^h$. It says that $e^h = 1+h+\text{terms that divided by h vanish as h tends to 0}$. – Mankind May 02 '15 at 19:04
  • does $\frac{e^h-1}{h}\approx 1$ imply that $e \approx (1+h)^{1/h}$? – John Joy May 02 '15 at 20:55
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    @JohnJoy: The result is true but the justification is not. $\frac{e^h-1}{h} \in 1+o(1)$ as $h \to 0$. This only gives $e^h \in 1+h+o(h)$ and hence $e \in (1+h+o(h))^\frac{1}{h}$. But nothing obviously shows that this implies what you want, which is $e \in (1+h)^\frac{1}{h}+o(1)$. It is true but not trivial that $e \in (1+h+o(h))^\frac{1}{h} \rightarrow e \in (1+h)^\frac{1}{h+o(h)} \subseteq (1+h)^\frac{1+o(1)}{h}$, but this is still nowhere near the desired result. – user21820 May 03 '15 at 04:13
  • I will add the link to some proofs of the limit you have used: Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ In any case, I would expect that this might be one of the limits students are expected to memorize. – Martin Sleziak May 03 '15 at 09:36
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$${\lim \limits_{h \to 0}} { {e^{2+h}-e^2 } \over {h} } = \frac{d}{dx}\left(e^x\right)\Big|_{x=2}.$$

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

abel
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