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I found a very tough limits question online. The question asks you to evaluate the limit $$\lim_{x \to 0}\frac{(x+4)^\frac{3}{2}+e^{x}-9}{x}$$ without using L'Hôpitals rule.

I tried to treat the top as a radical expression with the $e^x-9$ grouped and the other in root form to try to attempt rationalization. It did not work because you still get $\frac 0 0$.

I tried a trick of double rationalization but that did not work, got back to the starting. Second attempt I tried to let $x=z-4$, a substitution, but it still did not lead to something that could remove a zero from the numerator.

Then I tried to break this up into three fractions, by dividing $x$ into each term in the numerator, and I basically got $+\infty$, then can't do $e^x/x$ and then $-\infty$.

So I have exhausted all the algebraic tricks I can think of.

Anybody out there think they they can crack this one? Hope someone can.

Sincerely,

Palu

Cameron L. Williams
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Palu
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  • thanks for editing, I forgot to enclose in the dollar signs! – Palu Nov 01 '13 at 03:31
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    This is just the definition of the derivative of the numerator at $0$. – Ted Shifrin Nov 01 '13 at 03:31
  • Hi there, the question says: :"Use the definition of the derivative to find this limit" – Palu Nov 01 '13 at 03:35
  • So hence trying to model this against the definition of the derivative which is: lim h->0[f(x+h)-f(x)]/h. Is this what we need to do. Or do we do some complex manipulation to evaluate it. – Palu Nov 01 '13 at 03:38
  • So ted, are you saying that we set the numerator to zero, to try to see some relation. Palu – Palu Nov 01 '13 at 03:41
  • See the answer given by Zarrax. Your limit is precisely the definition of the derivative at $x=0$, so all you have to do is take the derivative and evaluate it at $0$. – Jaycob Coleman Nov 01 '13 at 03:44
  • See @Zarrax's solution. Even easier, set $f(x)=(x+4)^{3/2}+e^x$. – Ted Shifrin Nov 01 '13 at 03:45

6 Answers6

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Let $f(x) = (x+4)^\frac{3}{2}+e^{_{x}}-9$. Your limit can be written as $$\lim_{x \to 0}\frac{f(x)- f(0)}{x - 0}$$ Which is the definition of $f'(0)$. Thus the answer is ${3 \over 2}(0 + 4)^{1 \over 2} + e^0 = 4$.

Zarrax
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  • Hi Zarrax,I see now it, it is using what I call the secant form of the definition of the derivative. So you used this idea to find what f(x) is, but then you used power rule differentiation techniques and exponential differentiation. I am not sure if it is permitted to do it this way. BUT i could be wrong. Would one be able to figure this out if one was not allowed to have the derivative rules, and only by first principle formula. That is how i interpreted the question. – Palu Nov 01 '13 at 03:53
  • BUT Zarrax, i want to say that I love your solution! I have never seen a question like this before! Thanks for this solution! – Palu Nov 01 '13 at 03:53
  • Ok, so i guess it maybe the solution, just by saying its the definition of the derivative at zero. I will accept this as the answer, but I am still curious if this can be computed without L'Hopital and without using the derivative rules. – Palu Nov 01 '13 at 04:20
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    Well, you said no l'hopital but didn't say no calculus. Probably you could find the limit of $\frac{e^x - 1}{x}$ and $\frac{(x+4)^{3 \over 2} - 8}{x}$ straight from definitions and add. For the second one, you can do the multiplying by conjugate trick. For the first there is probably something using basic properties of $e^x$. – Zarrax Nov 01 '13 at 04:23
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    Ok, that is a good trick, I did not see this, break up the -9 to -1 and -8. Thanks!!! These are the kinds of computational tricks I like to learn! Really appreciate all your input Zarrax!!! Have a great day! – Palu Nov 01 '13 at 04:35
  • Do you have a formula for this definition? I couldn't get how $\lim_{x \to 0}\frac{f(x)- f(0)}{x - 0}$ gives the definition of $f'(0)$ – Abhishekstudent Sep 09 '16 at 15:39
  • This probably won't be allowed. This is kinda the same as L'Hospital's rule. – ShankRam Jan 13 '17 at 12:49
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This one is easily transformed into an algebraic limit. We need to make use of the following standard limit theorems $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ We can proceed in the following manner \begin{align} L &= \lim_{x \to 0}\frac{(x + 4)^{3/2} + e^{x} - 9}{x}\\ &= \lim_{x \to 0}\frac{(x + 4)^{3/2} - 8}{x} + \frac{e^{x} - 1}{x}\\ &= \lim_{x \to 0}\frac{(x + 4)^{3/2} - 8}{(x + 4) - 4} + 1\\ &= \lim_{t \to 4}\frac{t^{3/2} - 4^{3/2}}{t - 4} + 1\text{ (by putting }t = x + 4)\\ &= \frac{3}{2}\cdot 4^{1/2} + 1 = 4 \end{align}

Paramanand Singh
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f(x)=(〖(x+4)〗^(3/2)+e^x-9)/x

According to Maclauren's Series

f(x)=(x^0 f(0))/0!+(x^1 f^' (0))/1!+(x^2 f^'' (0))/2!+(x^3 f^''' (0))/3!+⋯

So, using this series We get

1) 〖(x+4)〗^(3/2)=8+3x+3/16 x^2+⋯

2) e^x=1+x+x^2/2+⋯

Substituting these in f(x) We'll get

f(x)=((8+3x+3/16 x^2+⋯)+(1+x+x^2/2+⋯)-9)/x

Combining terms of same powers we get

f(x)=((8+1-9)+(3+1)x+(3/16+1/2) x^2)/x Therefore, f(x)=4+8/16 x+⋯

The rest of the terms have higher degrees of x

So, now applying limits, we get

lim┬(x→0)⁡〖f(x)=lim┬(x→0)⁡〖(4+8/16 x+⋯〗 〗)=4

Yash Jain
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Just take 4 comman and use binomial expansion and then expand e^x and solve very simple Not much calculation

akshay
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Rewrite (x+4)^(3/2) as [4^(3/2) (1+x/4)^(3/2)] which is 8 (1+x/4)^(3/2). When x is small, (1+x/4)^(3/2) can be approximated by [1+(3/2)(x/4)] that is to say [1 + 3 x / 8]; then, close to zero, (x+4)^(3/2) is (8 + 3 x). On the other hand, when x is close to zero, Exp[x] can be approximated by (1 + x). Then, the numerator is approximated by : 8 + 3 x + 1 + x - 9 = 4 x. So the result is (4 x / x) = 4.

Claude Leibovici
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(x+4)^(3/2)-8/x+e^x-1/x (x+4)^(3/2)-8/x +1 (standard limit) ((x+4)^(3/2)-4^(3/2))((x+4)^(3/2)+4^(3/2))/x*((x+4)^(3/2)+4^(3/2)) +1 (x+4)^3-4^3/x*((x+4)^(3/2)+4^(3/2)) +1 put x=0 12*4/2*8 +1 4

Shravan Sridhar
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