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Disclaimer: I am a middle aged adult learning Calculus. This is not a student posting his homework assignment. Thank humanity for this great forum!

$$\lim_\limits{h\to 0}\frac{1}{h}\ln\left(\frac{2+h}{2}\right)$$

1) Can't directly sub the h. So, normally you reduce and cancel. Can you point me in the right direction? The directions say "manipulate the expression so L'Hopital's is used" I think L'hopital's is involved. Just not sure how to deal with the $\frac{1}{h}$

$$\lim_\limits{x\to \infty}\frac{x^k}{e^x}$$

2) Also, any tips on the one above? If $k$ is a positive integer, what is the limit above?

Thanks for any guidance.

JackOfAll
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5 Answers5

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If you must use l'Hospital's rule, then we can use it to note that:

$$\lim_{h\to 0}\frac{\ln(1+h/2)}{h}=\lim_{h\to0}\frac{1}{2}\frac{1}{1+h/2}$$

Direct substitution then gives us the answer:

$$\lim_{h\to0}\frac{1}{h}\ln\left(\frac{2+h}{2}\right) = \frac{1}{2}$$

For the second limit:

$$\lim_{x\to \infty}\frac{x^{k}}{e^{x}}$$

We note that we can write the Maclaurin series for $e^{x}$:

$$e^{x} = \sum_{k=0}^{\infty}\frac{x^{k}}{k!}$$

And so we can write the limit:

$$\lim_{x\to\infty}\frac{x^{k}}{\sum_{i=0}^{\infty}\frac{x^{i}}{i!}}$$

What does this mean for any integer value of $k$?

Thomas Russell
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$\ln(1+au)\sim_0 au$, hence $\dfrac1h\ln\Bigl(\dfrac{2+h}2\Bigr)=\dfrac1h\ln\Bigl(1+\dfrac h2\Bigr)\sim_0 \dfrac{\dfrac h2}{\,h\,}=\dfrac 12.$

Bernard
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  • What does the ~0 mean? – JackOfAll Mar 22 '16 at 12:08
  • It is how equivalence of functions in a neighbourhood of $0$ is denoted. Equivalence if often the easiest and shortest way to calculate limits. – Bernard Mar 22 '16 at 12:11
  • Is there another way? The directions say "manipulate the expression so L'Hopital's is used" – JackOfAll Mar 22 '16 at 12:21
  • You can use L'Hospital: just compute the limits of the quotient of the derivatives. In general, I don't recommend using L'Hospital as one must be careful in applying it. Furthermore it is equivalent to using Taylor's formula at order $1$. In the particular case, it is acceptable, though. – Bernard Mar 22 '16 at 13:09
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$$\frac{1}{h}\,\log\left(\frac{2+h}{2}\right) = \int_{0}^{1}\frac{dx}{2+hx}$$ hence by the dominated convergence theorem: $$ \lim_{h\to 0} \frac{1}{h}\,\log\left(\frac{2+h}{2}\right) = \int_{0}^{1}\frac{dx}{2}=\color{red}{\frac{1}{2}}.$$

Jack D'Aurizio
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Set $h=2n$ to find $$\dfrac12\cdot\lim_{n\to0}\dfrac{\ln(1+n)}n=?$$

See Determine the following limit as x approaches 0: $\frac{\ln(1+x)}x$

Community
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lab bhattacharjee
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  • Thanks. I tried your sub and got $\lim_\limits{2n\to 0}\frac{1}{2n}\ln\left(\frac{2+2n}{2}\right)$ = $\frac{1}{2}\lim_\limits{2n\to 0}\frac{1}{n}\ln(1+n))$ How does the limit go from 2n->0 to just n->0 ? – JackOfAll Mar 22 '16 at 12:04
  • Is there another way? The directions say "manipulate the expression so L'Hopital's is used" – JackOfAll Mar 22 '16 at 12:21
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For the first problem, we have $$\lim\limits_{h\to 0}\frac{1}{h}\ln\left(\frac{2+h}{2}\right)$$ $$=\lim\limits_{h\to 0}\frac{\ln\left(2+h\right)-\ln(2)}{h}$$ $$=\lim\limits_{h\to 0}\frac{\frac{\mathrm d}{\mathrm dh}\ln\left(2+h\right)-\frac{\mathrm d}{\mathrm dh}\ln(2)}{\frac{\mathrm d}{\mathrm dh}h}$$ $$=\lim\limits_{h\to 0}\frac{\frac{1}{2+h}-0}{1}=\lim\limits_{h\to 0}\frac{1}{2+h}=\frac12$$ For the second problem, for every $k\in\mathbb N$, we have $$\lim\limits_{x\to\infty}\frac{x^k}{e^x}=\lim\limits_{x\to\infty}\frac{\frac{\mathrm d^k}{\mathrm dx^k}x^k}{\frac{\mathrm d^k}{\mathrm dx^k}e^x}=\lim\limits_{x\to\infty}\frac{k!}{e^x}=0$$ Note that I simply applied L'Hôpital's rule to the original limit $k$ number of times.

k170
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