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I am wondering about:

$$ \lim_{x \rightarrow -1} \frac{x+1}{ \sqrt{x+5}-2 } $$

which seems to be an asymptotic function with each side of its corresponding graph approaching negative or positive infinity at the value of -1.

Solved by WolframAlpha, the expected result was calculated: Click to see result.

So, I was just learning for a math exam and using the website KhanAcademy to refresh my math skills, where above term was to be solved, and indeed a two-sided limit should be found, which apparently seems to be 4.

This was done by first rationalizing the term in this manner, which results in: $$ \lim_{x \rightarrow -1}( \sqrt{x+5}+2)\text{ for }x \neq -1 $$

Substituting x for -1 in this simplified term doesn't result in getting $\frac{0}{0}$ but 4, which apparently means that 4 is the both-sided limit (even though the root of 4 has 2 as well as -2 as the possible solution, making 0 another viable limit?). I can't proof which one of those two approaches contains an error and therefore wanted to ask here, I hope someone can clarify this matter :)

Best regards

Thomas Andrews
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Reisbrot
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4 Answers4

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The limit is $4$, in fact: $$ \lim_{x \to -1}\frac{x+1}{ \sqrt{x+5}-2 }=\lim_{x \to -1} \frac{x+1}{ \sqrt{x+5}-2 }\frac{\sqrt{x+5}+2}{\sqrt{x+5}+2}=\lim_{x \to -1}\frac{(1+x)(\sqrt{x+5}+2)}{x+1} $$ The last limit is equal to: $$\lim_{x \to -1}\frac{(1+x)(\sqrt{x+5}+2)}{x+1}=\lim_{x \to -1}(\sqrt{x+5}+2)=2+2=4 $$

Sine of the Time
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No need of l'Hôpital here. Considering the function $$g(x)=\sqrt{x+5},\quad g'(x)=\frac1{2\sqrt{x+5}},$$ just notice that $$\lim_{x\to-1}\frac{\sqrt{x+5}-2}{x+1}=\lim_{x\to-1}\frac{g(x)-g(-1)}{x-(-1)}=g'(-1)=\frac14,$$ hence your limit is the inverse, i.e. $4.$

Anne Bauval
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You computed the wrong limit in Wolfram:

$$4=\lim_{x \to -1}\frac{x+1}{ \sqrt{x+5}-2 } \neq \lim_{x \to -1}\frac{1}{ \sqrt{x+5}-2 }$$

Weierstraß Ramirez
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Another way to approach it (according to the L'Hôpital rule): \begin{align*} \lim_{x\to-1}\frac{x + 1}{\sqrt{x + 5} - 2} & = \lim_{x\to-1}\frac{1}{\frac{1}{2\sqrt{x + 5}}} = \lim_{x\to-1}2\sqrt{x + 5} = 4 \end{align*}

Hopefully this helps!

Átila Correia
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