$\lim_{x\to0}\frac{\sqrt{5x+3}-\sqrt 3}{5^{\sin(7x)}-1}$ find without using l'Hôpital's rule. 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.
 A: 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?
A: 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 $\dfrac{A}{B}$.
A: 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.
