How to solve $\displaystyle\lim_{x\to 0}\tfrac{\sqrt{x+25}-5} {\sqrt{x+16}-4}$ \begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4}
\end{eqnarray}
Undefined limit
\begin{eqnarray}
\frac{0} {0}
\end{eqnarray}
 A: METHOD 1:
One quick way is to write
$$\sqrt{x+a^2}=|a|\sqrt{1+x/a^2}=|a|\,\left(1+\frac12 \left(\frac{x}{a^2}\right)+O\left(x\right)^2\right)$$
Then, we have
$$\frac{\sqrt{x+25}-5}{\sqrt{x+16}-4}=\frac45 +O(x)\to \frac45$$

METHOD 2:
Using L'Hospital's Rule gives
$$\lim_{x\to 0}\frac{\sqrt{x+25}-5}{\sqrt{x+16}-4}=\lim_{x\to 0}\frac{\frac{1}{2\sqrt{x+25}}}{\frac{1}{2\sqrt{x+16}}}=\frac45$$
A: Hint: By multiplying the numerator and denominator by $(\sqrt{x+25}+5)(\sqrt{x+16}+4)$, and then using the difference of squares identity $(a-b)(a+b) = a^2-b^2$, we have:
$\dfrac{\sqrt{x+25}-5}{\sqrt{x+16}-4}$ $= \dfrac{(\sqrt{x+25}-5)(\sqrt{x+25}+5)(\sqrt{x+16}+4)}{(\sqrt{x+16}-4)(\sqrt{x+16}+4)(\sqrt{x+25}+5)}$ $= \dfrac{x(\sqrt{x+16}+4)}{x(\sqrt{x+25}+5)}$.
Can you take the limit of this expression as $x \to 0$?
A: Note that, 
$$
\lim_{x \to 0}\dfrac{\sqrt{x + a^2} - a}{x} = \lim_{x \to 0}\dfrac{1}{\sqrt{x + a^2} + a} = \dfrac{1}{2a} 
$$
for $a > 0$. Thus,
\begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4} = \dfrac{\lim_{x \to 0}\dfrac{\sqrt{x + 25} - 5}{x}}{\lim_{x \to 0}\dfrac{\sqrt{x + 16} - 4}{x}} = \dfrac{1/10}{1/8} = \dfrac{4}{5}
\end{eqnarray}
A: Let $f(x) = \sqrt {x+25}, g(x) = \sqrt {x+16}.$ We are looking at
$$\frac{f(x) - f(0)}{g(x)-g(0)}= \frac{(f(x) - f(0))/x}{(g(x)-g(0))/x}.$$
By the definition of the derivative, the numerator $\to f'(0),$ the denominator $\to g'(0).$ So the desired limit is $f'(0)/g'(0),$ and now we have an easy computation. (Note: We are not using L'Hopital here.)
A: Given $\displaystyle \lim_{x\to 0}\frac{\sqrt{x+25}-5}{\sqrt{x+16}-4} \times \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4} \times \frac{\sqrt{x+25}+5}{\sqrt{x+25}+5}$
We Get $\displaystyle \lim_{x\rightarrow 0} \frac{x}{x} \times \frac{\sqrt{x+16}+4}{\sqrt{x+25}+5} = \frac{8}{10}$
