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How do I find $$\lim_{x\to 0}\frac{(x+4)^{3/2}+e^x-9}{x}$$ without l'hôpital rule? I know from l'hôpital the answer is 4.

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    $\begingroup$ I've tried conjugates because the x+4 equation is technically a sqaure root but that didn't work. $\endgroup$ – Umeko Inoue May 17 '17 at 0:34
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    $\begingroup$ Apart from that I've been googling and trying to find methods (and desipher the hint suggested below) $\endgroup$ – Umeko Inoue May 17 '17 at 0:40
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    $\begingroup$ To help explain the hint, $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)$, by the definition of derivative. $\endgroup$ – vadim123 May 17 '17 at 3:25
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Hint: The expression equals

$$ \frac{f(x) - f(0) + g(x) - g(0)}{x-0}$$

if you choose $f,g$ appropriately.

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  • $\begingroup$ That's the approach I would have suggested. $\endgroup$ – Mark Viola May 17 '17 at 1:12
  • $\begingroup$ Is this not literally l'Hopital's rule with the added convenience that the function in the denominator happens to be $x$ rather than some $h(x)$? $\endgroup$ – LLlAMnYP May 17 '17 at 7:19
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    $\begingroup$ @LLlAMnYP: No L'Hospital's Rule works by differentiating and then taking a limit. Here it is just shown that the limit in question is a derivative namely $(f - g)'(0)$ for suitable $f, g$. Recognizing a limit as a derivative and recognizing a $0/0$ indeterminate form and further applying L'Hospital on it are two different things and first one is simpler. In general using L'Hospital's Rule to evaluate limits of type $\lim_{x \to a}\dfrac{f(x) - f(a)}{x - a}$ is a bad idea and it may not work all the time. $\endgroup$ – Paramanand Singh May 17 '17 at 11:05
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    $\begingroup$ @LLlAMnYP Look at it this way: The approach I'm suggesting would be feasible in the first few weeks of basic calculus, even before the mean value theorem. LHR comes later, usually around the time Taylor's theorem is discussed, and depends on the fancier generalized MVT. Also as our friend Paramanand Singh notes, all we need to use this approach is that both $f'(0),g'(0)$ exist - we do not need to know $f'(x),g'(x)$ exist for any other $x$ as in LHR. $\endgroup$ – zhw. May 17 '17 at 23:58
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The limit can be found using neither Taylor expansions, nor derivatives, but with much more basic approaches.

First, a consequence of the well-known limit

$$ \lim_{x\rightarrow0} \left(1 +\frac{1}{x}\right)^x = e$$

is

$$ \lim_{x\rightarrow0} \frac{e^x -1}{x} = 1 $$

so the limit in OP is equal to

$$ 1 + \lim_{x\rightarrow0} \frac{(x+4)^{3/2} -8}{x} $$

so long as the second term exists. We calculate the second term as

$$ \lim_{x\rightarrow0} \frac{((x+4)^{3/2}-8)((x+4)^{3/2}+8)}{x((x+4)^{3/2}+8)} = \lim_{x\rightarrow0} \frac{(x+4)^3-64}{x((x+4)^{3/2}+8)} = \\ \lim_{x\rightarrow0} \frac{(x+4)^3-64}{x} \lim_{x\rightarrow0} \frac{1}{((x+4)^{3/2}+8)} = \frac{1}{16} \lim_{x\rightarrow0}\frac{48x+12x^2+x^3}{x}=\frac{48}{16}=3$$

where again then transition from the first line to the second is valid so long as both limits exist (which, as you see, they do).

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  • $\begingroup$ Taylor's theorem is independent of L'Hospital's Rule. There is a proof using L'Hospital's Rule, but it does not make them equivalent. There are other proofs too. $\endgroup$ – Paramanand Singh May 17 '17 at 11:08
  • $\begingroup$ @ParamanandSingh you're more than likely right. This problem took me back to a calculus exam with a simple limit $\lim_{x\rightarrow0}\frac{\sin 5x}{\sin 7x}$ to be done without l'Hopital. Attempts to do this using Taylor were also for some reason considered implicitly out of bounds. So we had to start from $\lim_{x\rightarrow0}\frac{\sin x}{x}$ $\endgroup$ – LLlAMnYP May 18 '17 at 6:04
  • $\begingroup$ That particular limit can be done just by using the formulas for $\sin 5x,\sin 7x$ in terms of $\sin x$. Note that if $n$ is odd then $\sin nx=n\sin x+\dots$ and $\sin x$ can be taken out as factor. Canceling it we get the limit $5/7$. Also see this answer math.stackexchange.com/a/2101475/72031 $\endgroup$ – Paramanand Singh May 18 '17 at 6:11
  • $\begingroup$ @ParamanandSingh sure. At that exam the correct solution for that particular question was $\frac{\sin 5x}{\sin 7x} = \frac{5}{7}\frac{\sin 5x}{5x}\frac{7x}{\sin 7x}$ $\endgroup$ – LLlAMnYP May 18 '17 at 9:57
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    $\begingroup$ My answer is not close to LHR, so please stop saying it is. It will do you well to understand this point. $\endgroup$ – zhw. May 21 '17 at 1:09
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Less elegant than zhw'x answer.

For $(x+4)^{3/2}$, use the generalized binomial theorem of Taylor expansion $$(x+4)^{3/2}=8+3 x+\frac{3 x^2}{16}+O\left(x^3\right)$$ Using the standard Taylor expnasion of $e^x$, we then have $$(x+4)^{3/2}+e^x-9=\left(8+3 x+\frac{3 x^2}{16}+O\left(x^3\right) \right)+\left( 1+x+\frac{x^2}{2}+O\left(x^3\right)\right)-9$$ $$(x+4)^{3/2}+e^x-9=4 x+\frac{11 }{16}x^2+O\left(x^3\right)$$ $$\frac{(x+4)^{3/2}+e^x-9}{x}=4 +\frac{11 }{16}x+O\left(x^2\right)$$ which shows the limit and how it is approached.

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