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I have to find this limit without using l'Hôpital's rule:

$$\lim_{x\to0}\frac{\sqrt{5x+3}-\sqrt 3}{5^{\sin(7x)}-1}$$

I have no idea how to do it. L'Hôpital's rule makes it easy, but how should I go about calculating this without using the rule?

I should add that I'm not supposed to use anything but the most basic methods and facts. I was thinking I should use the squeeze theorem, but I'm not seeing any estimates that work.

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Are you allowed to use the definition of the derivative, $f'(x)=\lim_{h\to0}(f(x+h)-f(x))/h$? –  Gerry Myerson Dec 11 '12 at 0:28
    
@GerryMyerson Well, I don't know for sure. If I were, I could just imitate the proof of l'Hôpital's rule, and I think this would be cheating. But I don't really know because it's a problem I've been given by a girl who asked me to help her with her math. I don't attend that course. –  Bartek Dec 11 '12 at 0:31
    
Additionally do you have any information about the growth of functions, and can you find bounds (upper and lower) on sin(7x) that might be helpful with squeeze theorem or a substitution (think about the idea of $a^{\sin(x)}$ when x = $\frac{\pi}{2}$, $0$, and $\frac{3\pi}{2}$)? –  Vilid Dec 11 '12 at 0:35
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3 Answers

up vote 4 down vote accepted

You will need to use the following identities: $$\lim_{x\to 0}\frac{e^x-1}{x}=1,\hspace{10pt} \lim_{x\to 0}\frac{\sin x}{x}=1$$ Then $5^{\sin(7x)}=e^{\sin(7x)\ln5}$, so $$\lim_{x\to 0}\frac{5^{\sin(7x)}-1}{\sin(7x)\ln5}=\lim_{t\to 0}\frac{e^{t}-1}{t}=1$$ Using change of variable $t=\sin(7x)\ln5$.
Hence: $$\begin{align*}\lim_{x\to0}\frac{\sqrt{5x+3}-\sqrt 3}{5^{\sin(7x)}-1}&=\lim_{x\to0}\frac{\sqrt{5x+3}-\sqrt 3}{5^{\sin(7x)}-1}\frac{\sqrt{5x+3}+\sqrt 3}{\sqrt{5x+3}+\sqrt 3}\frac{\sin(7x)\ln5}{\sin(7x)\ln5}\\ &=\lim_{x\to0}\frac{\sin(7x)\ln5}{5^{\sin(7x)}-1}\frac{1}{\sqrt{5x+3}+\sqrt 3}\frac{7x}{\sin(7x)}\frac{5}{7\ln5}\end{align*}$$ Can you continue?

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Yes, I can of course. Thank you very much! I will have to think about how to prove the first identity without using l'Hôpital's rule, but I think the squeeze theorem should work here just fine, right? (I know how to prove the second one using the squeeze theorem.) –  Bartek Dec 11 '12 at 0:57
    
You don't need l'Hôpital's rule for any of those two. First show that $\lim_{x\to 0}\left(1+x\right)^{\frac1x}=e$, then use it to show that $\lim_{x\to 0}\frac{\ln(1+x)}x=1$ and, finally, deduce the first limit. –  Dennis Gulko Dec 11 '12 at 1:07
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Hint: As for the nominator, standard $(a+b)(a-b) = a^2-b^2$ trick should do the job. As for the denominator, you can expand $5^{\sin 7x}$ into its Taylor series (the linear term should be enough) and bound it from both sides and use this lemma.

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Multiply top and bottom by $x$.

We first deal with $$\lim_{x\to 0}\frac{\sqrt{5x+3}-\sqrt{3}}{x}.$$ If the $x$ were an $h$, we would recognize that the limit is by definition the derivative of $f(t)=\sqrt{5t+3}$ at $t=0$. So calculate the derivative, using differentiation formulas. Evaluate the derivative at $t=0$, and call the result $A$.

It remains to find $$\lim_{x\to 0}\frac{5^{\sin 7x}-1}{x}.$$ Again, by definition, this is the derivative of the function $5^{\sin 7t}$ at $t=0$. Calculate the derivative at $0$, using $5^y=e^{y\log 5}$. Let $B$ be the value of the derivative at $t=0$.

Then the answer to our limit problem is $\dfrac{A}{B}$.

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