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Please evaluate the following limit for me:

$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$

I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the denominator so it always stay in the indeterminate form.

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    $\begingroup$ You should have gotten $x^2 + 8 - (3^2) = x^2 - 1 = (x+1)(x-1)$ in the numerator, not simply $x^2$ $\endgroup$
    – amWhy
    Commented Sep 21, 2014 at 16:21

7 Answers 7

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One way to solve the problem is to consider the function $$f(x)=\sqrt{x^2+8}$$ so the desired limit is

$$\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=f'(-1)=\frac{x}{\sqrt{x^2+8}}\Bigg|_{x=-1}=-\frac13$$

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  • $\begingroup$ Never thought of this way to solve this problem! $\endgroup$ Commented Sep 21, 2014 at 21:34
  • $\begingroup$ How did you pass from the limit to the derivative? $\endgroup$
    – seldon
    Commented Sep 22, 2014 at 14:33
  • $\begingroup$ @mattecapu The derivative of $f$ evaluated at $x_0$ is defined as: $$f^\prime(x_0):=\lim_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}.$$ In this problem one may recognize that $x_0=-1$ and $f:x\mapsto\sqrt{x^2+8}$. $\endgroup$
    – Hakim
    Commented Sep 22, 2014 at 15:49
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Multiply the function by $$\frac{\sqrt{x^2 + 8} +3}{\sqrt{x^2 + 8}+3}$$

You'll have a difference of squares in the numerator of the form $(a -b)(a+ b)$ which, of course, is $a^2 - b^2$. You should have gotten, in the numerator: $$(\sqrt{x^2 + 8})^2 - (3^2) = x^2 + 8 - 9 = x^2 - 1 = (x+1)(x-1)$$

Simplifying, you'll get $$\lim_{x\to -1} \frac {\overbrace{x^2 -1}^{(x + 1)(x-1)}}{(x+1)(\sqrt{x^2 + 8} +3)} = \lim_{x\to -1} \frac {x-1}{\sqrt{x^2 + 8} +3} = \frac{-2}{6} = -\frac 13$$

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Use L'hopital's rule (Valid since the function is of the form $\frac{0}{0}$). The derivative of the top has limt equal to $-\frac{1}{3}$ and the derivative of the denominator is the constant $1$.

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Multiplying $$\frac{\sqrt{x^2+8}-3}{x+1}$$ by $$\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}\ \ (=1)$$ gives you $$\begin{align}\lim_{x\to -1}\frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)}&=\lim_{x\to -1}\frac{\color{red}{(x+1)}(x-1)}{\color{red}{(x+1)}(\sqrt{x^2+8}+3)}\\&=\lim_{x\to -1}\frac{x-1}{\sqrt{x^2+8}+3}\\&=\frac{-2}{\sqrt 9+3}\\& =-\frac 13.\end{align}$$

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we have $\frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)}$ simplifying this we get $\lim_{x \to -1}\frac{x-1}{\sqrt{x^2+8}+3}$
Sonnhard.

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$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} = \lim_{x \to -1} \frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)} \\ \\ = \lim_{x \to -1} \frac{x^2+8-9}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x^2-1}{(x+1)(\sqrt{x^2+8}+3)}=\\ \\ \lim_{x \to -1} \frac{(x-1)(x+1)}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x-1}{\sqrt{x^2+8}+3}=\frac{-2}{6}=-\frac{1}{3}$$

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Problem: $$\lim_{x\to-1}\frac{\sqrt{x^2+8}-3}{x+1}$$ Rationalize the numerator by multiplying by its conjugate $$=\lim_{x\to-1}\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$ Multiply out the numerator $$=\lim_{x\to-1}\frac{\sqrt{x^2+8}^2+\boxed{3\sqrt{x^2+8}-3\sqrt{x^2+8}}+-3*3}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$ Simplify $$=\lim_{x\to-1}\frac{x^2+8-9}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$ $$=\lim_{x\to-1}\frac{x^2-1}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$ Factor the numerator $$=\lim_{x\to-1}\frac{\left(x-1\right)\boxed{\left(x+1\right)}}{\boxed{\left(x+1\right)}\left(\sqrt{x^2+8}+3\right)}$$ Cancel $$=\lim_{x\to-1}\frac{x-1}{\sqrt{x^2+8}+3}$$ Plug in $-1$ $$=\frac{\left(-1\right)-1}{\sqrt{\left(-1\right)^2+8}+3}$$ Simplify $$=-\frac{1}{3}$$

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  • $\begingroup$ Thank you. I appreciate your valuable answer and your precious time. :) $\endgroup$
    – offchan
    Commented Oct 16, 2014 at 19:09

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