Limit using definition of $e$ I'm trying to calculate the following limit:
$$
\lim_{x\to0}\bigl(-4x + \sqrt{16x^2 + 1}\,\bigr)^{b/x}
$$
My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $\exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.
Does anyone have any advice?
 A: You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=\exp \frac{b}{x}(-4x+o(x))=\exp (-4b+o(1))=\exp(-4b)+o(1).$$
A: Letting $L = \lim_{X \to 0} (-4x+\sqrt{16x^2+1})^{\frac{b}{x}}$ we have;
$$\ln{L}=b \cdot \lim_{X \to 0}  \frac{\ln{-4x+\sqrt{16x^2+1}}}{x}  $$
By using L'Hôpital's Rule, this is equal to:
$$\ln{L}=b \cdot \lim_{X \to 0}  \frac{-4+\frac{32x}{2 \cdot \sqrt{16x^2+1}}}{-4x+\sqrt{16x^2+1}}  $$
$$=-4 \cdot b$$
$$\therefore L=e^{-4 \cdot b}$$
A: It's basically right:
$$
-4x+\sqrt{1+16x^2}=-4x+1+\frac{16x^2}{2}+o(x^2)=1-4x+o(x)
$$
Now you can substitute $-4x=y$, so the limit becomes
$$
\lim_{y\to0}\bigl((1+y+o(y))^{1/y}\bigr)^{-4b}
$$
and $\lim_{y\to0}(1+y)^{1/y}=e$ is known.
A: Consider $$y=\bigr( \sqrt{16x^2 + 1}-4x\bigr)^{\frac bx}$$ By Taylor or binomial expansion, we have
$$\sqrt{16x^2 + 1}=1+8 x^2+O\left(x^4\right)$$ that is to say
$$\sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+O\left(x^4\right)$$ Now
$$\log(y)=\frac bx \log\bigr( \sqrt{16x^2 + 1}-4x\bigr)=\frac bx\log\bigr(1-4 x+8 x^2+O\left(x^4\right)\bigr)=\frac bx\bigr( -4 x+\frac{32 x^3}{3}+O\left(x^4\right)\bigr)$$
$$\log(y)=-4 b+\frac{32 b x^2}{3}+O\left(x^3\right)$$ Continue with Taylor
$$y=e^{\log(y)}=e^{-4 b}+\frac{32}{3} b e^{-4 b} x^2+O\left(x^3\right)$$ which shows the limit and how it is approached.
A: Here is a way by "enforcing" a $\color{red}{1}$ in the brackets and using the standard limit for $e$:


*

*$y_n \stackrel{n \to \infty}{\longrightarrow}0 \Rightarrow (1+y_n)^{\frac{1}{y_n}}\stackrel{n \to \infty}{\longrightarrow}e$
\begin{eqnarray*}
\bigl(-4x + \sqrt{16x^2 + 1}\,\bigr)^{b/x} & = & \left(\color{red}{1} + \left(\color{blue}{-4x + \sqrt{16x^2+1} -1}\right) \right)^{b/x}\\
& = & \left(\underbrace{\left(\color{red}{1} + \left(\color{blue}{-4x + \sqrt{16x^2+1} -1}\right) \right)^{\frac{1}{\color{blue}{-4x + \sqrt{16x^2+1} -1}}}}_{\stackrel{x\to 0}{\rightarrow}e} \right)^{b\cdot \underbrace{\frac{\color{blue}{-4x + \sqrt{16x^2+1} -1}}{x}}_{\stackrel{x\to 0}{\rightarrow}-4=f'(0),\; f(x) = -4x + \sqrt{16x^2+1}}} \\
& \stackrel{x\to 0}{\longrightarrow} & e^{b\cdot (-4)}
\end{eqnarray*}
