Jtdylktuy

I'm trying to calculate the following limit:
$$
lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
$$
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?

You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$

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:
$$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
$$=-4 cdot b$$
$$therefore L=e^{-4 cdot b}$$

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
$$
lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
$$
and $lim_{yto0}(1+y)^{1/y}=e$ is known.

Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
$$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
$$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
$$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
$$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
$$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.

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{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
  & stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
  end{eqnarray*}