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Is there a way to evaluate the following limit without using L'Hopital's rule? $$\lim_{t \to 0} \frac{e^{5t} -1} {t}$$

Based on this answer here I'm guessing that the series expansion for $e^{5t}$ would be helpful, but I'm not sure how to properly use it (I don't quite understand the rationale for the way the inequalities are setup).

Aside from series expansion, are there any other methods?

Rodrigo de Azevedo
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Slecker
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  • $$\lim_{t \to 0} \frac{e^{5t} -e^{5 \cdot 0}}{t - 0 }$$ – rlartiga Oct 02 '20 at 15:57
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    Can you use the known limit $\displaystyle \lim_{t \to 0} \frac {e^t-1}t = 1$? – player3236 Oct 02 '20 at 15:57
  • @player3236 Not sure how to go about doing so. Closest I get is $\lim_{t \to 0} \frac {(e^t)^5 - 1} {t}$. Not sure how to proceed from there, or if I'm even on the right track. – Slecker Oct 02 '20 at 16:31
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    Simply use the fact that $\displaystyle \lim_{t\to0} \frac {e^{5t}-1}{5t}$ is also $1$. – player3236 Oct 02 '20 at 16:32
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    @player3236 Ahhh I see, thank you! So $\lim_{t \to 0} \frac {e^{5t} - 1} {t} \cdot \frac {5}{5} = \lim_{t \to 0} 5\cdot\frac {e^{5t} - 1} {5t} = 5\cdot\lim_{t \to 0} \frac {e^{5t} - 1} {5t} = 5$. – Slecker Oct 02 '20 at 16:39
  • @player3236 As a follow-up question, what method could I use to verify $\lim_{t \to 0} \frac {e^t -1} {t} = 1$? It's not a limit I've seen until now. – Slecker Oct 02 '20 at 16:41
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    Huh. I asked if you know this limit before because this is one of the first limits I've learnt, along with $\displaystyle \lim_{x \to 0} \frac {\ln (1 + x)}x = 1$. You can find some proofs here. – player3236 Oct 02 '20 at 16:48
  • @player3236 I'm using James Stewart's Calculus (8th edition) to self-study. Perhaps those limits are somewhere in here and I've just overlooked them. Anyway, thank you! – Slecker Oct 02 '20 at 16:53

3 Answers3

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If you use $f(x)=e^{5x}$ then $f'(x)=5e^{5x}$

$$\lim_{t \to 0} \frac{e^{5t} - e^{5 \cdot 0}} {t-0} = \lim_{t \to 0} \frac{f(t) - f(0)} {t-0}=f'(0)=5e^{5\cdot 0}=5$$

rlartiga
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You also can use the series expansion $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\cdot\cdot\cdot$, so we have $$e^{5t}=\sum_{n=0}^{\infty}\frac{(5t)^n}{n!}=1+5t+\frac{(5t)^2}{2!}+\cdot\cdot\cdot$$

Thus $$\frac{e^{5t}-1}{t}=5+\frac{5^2}{2!}t+\cdot\cdot\cdot$$

Now take the limit.

Alessio K
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  • Ah thank you! I actually did the same on paper, but I wasn't sure if I was substituting in the series expansion for $e^{5t}$ correctly because the answer I linked was using inequalities when substituting it in. – Slecker Oct 02 '20 at 16:54
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    You're welcome! In the answer from the link it was $e^{-\frac{1}{t}}$ which is not continuous at $0$. Here you have $e^{5t}$ which is simpler to work with and just substitute into the series $$e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ (as $x=5t$) which is valid for all $x\in\mathbb R$. – Alessio K Oct 02 '20 at 17:02
  • Ahh, I see. Just to clarify, in general, if a function is continuous at the point in question (the value being approached), then we can simply substitute in its series expansion. Whereas if a function is discontinuous at the value being approached, then other conditions have to be considered when substituting, such as the inequalities in the answer I linked. Is this a correct notion? – Slecker Oct 02 '20 at 17:13
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    Yes you are correct. Note that the limit does not always exist. See this for example. – Alessio K Oct 02 '20 at 17:23
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    Great, thank you for all your help! :) – Slecker Oct 02 '20 at 17:32
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Note that, $$\lim_{t\to 0}\dfrac{e^{5t} - 1}{t} = \lim_{5t\to 0}\dfrac{5(e^{5t} - 1)}{5t}$$ Here you can apply direct result that, $\lim_{x\to0} \dfrac{e^x-1}{x} = 1$ $$\implies 5(1) = 5$$