2

Given the limit: $$ \lim\limits_{x\to 1}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}\right) = \alpha $$

Find the value of $\alpha$

Could not get my head around on how to simplify the nominator.

Anyone feeling like this should be easy?

Thanks in advance.

E.Z
  • 273

4 Answers4

3

Let se $x=1+y$ with $y\to0$ then

$$\lim\limits_{x\to 1}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}\right) = \lim\limits_{y\to 0}\left(\frac{\sqrt{y^2+4y+8-8\sqrt{1+y}}}{\log(1+y)}\right)$$

and note that

  • $\sqrt{1+y}=1+\frac12 y-\frac18y^2 +o(y^2)$
  • $\log(1+y)=y+o(y)$

then

$$\frac{\sqrt{y^2+4y+8-8\sqrt{1+y}}}{\log(1+y)}=\frac{\sqrt{y^2+4y+8-8-4y+y^2+o(y^2)}}{y+o(y)}=\frac{\sqrt{2y^2+o(y^2)}}{y+o(y)}=\frac{|y|\sqrt{2+o(1)}}{y+o(y)}=\frac{|y|}{y}\frac{\sqrt{2+o(1)}}{{1+o(1)}}$$

thus the limit doesn't exist.

user
  • 154,566
2

As regards the numerator, we can rewrite it in the following way $$x^2+2x+5-8\sqrt{x}=(x+2\sqrt{x}+5)(x-2\sqrt{x}+1)=(x+2\sqrt{x}+5)(\sqrt{x}-1)^2.$$ Moreover, $\log(x)=2\log(1+(\sqrt{x}-1))$. Therefore $$\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\log(x)}= \frac{\sqrt{x+2\sqrt{x}+5}\cdot |\sqrt{x}-1|}{2\log(1+(\sqrt{x}-1))}.$$ Now recall that $\lim_{t\to 0}\frac{\log(1+t)}{t}=1$, and consider separately $x\to 1^+$ and $x\to 1^-$ (the given limit for $x\to 1$ does not exist!).

Can you take it from here?

Robert Z
  • 145,942
2

Hint: Limit does not exist as $x \to 1^- ln(x) <0$ and $x \to 1^+ ln(x) >0$ and numerator is positive. In this case as pointed out in comments by King Tut if R.H.L. comes to be 0 then limit exist.
For finding R.H.L.
$$\lim\limits_{x\to 1^+}\left(\frac{\sqrt{x^2+2x+5-8\sqrt{x}}}{\sqrt{\log^2(x)}}\right) = \alpha \implies \lim\limits_{x\to 1^+}\left(\frac{{x^2+2x+5-8\sqrt{x}}}{\log^2(x)}\right) = \alpha^2$$ Now apply L-Hospital rule.

kayush
  • 2,441
  • 1
    Unfortunately your first statement is not good. The limit could have been zero in which case expression being negative or positive in left side and right side wouldnt matter. – King Tut Apr 07 '18 at 08:03
  • @KingTut you are correct! I have edited my post. thanks for pointing out. – kayush Apr 07 '18 at 08:05
2

To simplifiy, try to use $x-1 = t$ and $t\to 0$ in the limit:

$$\lim_{t\to 0} \frac{\sqrt{(1+t)^2+2(1+t)+5-8\sqrt{1+t}}}{\ln(1+t)}\\ =\lim_{t\to 0} \frac{\sqrt{t^2+4t+8(1-\sqrt{1+t})}}{\ln(1+t)}\\ = \lim_{t\to 0} \frac{\sqrt{t^2+4t-8\tfrac{t}{1+\sqrt{1+t}}}}{\ln(1+t)}$$

Now divide by $t$ on both numerator and denominator and rationalise, further use standard limits:

For $t\to 0^+$ we can take $t$ = $\sqrt{t^2}$

$$ \lim_{t\to 0} \frac{\sqrt{1+\tfrac{4(\sqrt{1+t}-1)}{t(1+\sqrt{1+t})}}}{\tfrac{\ln(1+t)}{t}} = \lim_{t\to 0} \frac{\sqrt{1+\tfrac{4t}{t(1+\sqrt{1+t})^2}}}{\tfrac{\ln(1+t)}{t}} = \sqrt{1+1} = \sqrt{2}$$

to get the limit as $\sqrt{2}$

But for $t\to 0^-$, $\sqrt{t^2} = -t$, so the limit will be $-\sqrt 2$

Thus $LHL \neq RHL$ and limit does not exist.

King Tut
  • 3,074