# Solving limit without L'Hôpital

I need to solve this limit without L'Hôpital's rule. These questions always seem to have some algebraic trick which I just can't see this time.

$$\lim_{x\to0} \frac{5-\sqrt{x+25}}{x}$$

Could someone give me a hint as to what I need to do to the fraction to make this work? Thanks!

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I'm not sure why anyone would ask you to solve a limit with an easy solution, using a difficult solution. The instructor needs to find examples that require the skills expected for this limit, whilst L'Hopital's is impossible or impractical. If such a case doesn't exist, then you don't need the skills it teaches ;) – Cruncher Apr 25 '14 at 15:26
@Cruncher: To evaluate such a limit by L'Hôpital, you need to know that $\frac{d}{dx}\sqrt x = 1/(2\sqrt x)$, and to prove that formula correct (from the definition of derivative), you need to be able to evaluate this kind of limit. So refraining from using L'H is not some artificial restriction imposed by the teacher because they couldn't think of any other way to test the student's knowledge of a particular "skill"; it's a restriction which arises out of logical considerations in the structure of the foundations of the theory. (A similar case: $\lim_{x\to 0}\frac{\sin x}{x}$.) – user21467 Apr 25 '14 at 21:15
@StevenTaschuk then the proof of $\frac{d}{dx}\sqrt{x}$ should be done separately. To impose such a restriction means they've likely already learned l'hopitale, and learned how to differentiate roots. So if the teacher didn't take "logical consideration in the structure of the foundations of the theory" then, then why start now? (tldr: Give these kinds of questions before teaching l'hopitale) – Cruncher Apr 26 '14 at 12:56
@Cruncher - Pedagogical order is often different from logical order. It can make sense to teach mid-level techniques first and supply the logical foundations later; for one thing, this can bring interesting applications within reach quickly, which provides motivation for the foundational work. This is a perfectly sound pedagogical approach, assuming (as I do) that the students are capable of understanding that the order they encountered something in might be different from the order that it's built in. – user21467 Apr 26 '14 at 14:08

$$\lim_{x\to0} \frac{5-\sqrt{x+25}}{x}=\lim_{x\to0} \frac{(5-\sqrt{x+25)}(5+\sqrt{x+25})}{x(5+\sqrt{x+25})}=\lim_{x\to0} \frac{25-(x+25)}{x(5+\sqrt{x+25})}=-\frac{1}{10}$$

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I fixed typo's (2 parentheses were missing). Cheers. – Claude Leibovici Apr 25 '14 at 4:16
thanks sir, my regards to you. – ketan Apr 25 '14 at 4:18
You are very welcome ! – Claude Leibovici Apr 25 '14 at 4:19

Let $t^2=x+25$, then $t=\sqrt{x+25}.$ Then we have $$\lim_{t\to 5}\frac{5-t}{t^2-25}=\lim_{t\to 5}\dfrac{5-t}{(t+5)(t-5)}=-\lim_{t\to 5}\dfrac{5-t}{(t+5)(5-t)}=-\lim_{t\to 5}\dfrac{1}{t+5}=-\dfrac{1}{10}.$$

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This answer is most creative of all of them,I am simply loving it – Vanio Begic Apr 25 '14 at 4:38
Choosing the negative root is equally valid, and soon simplifies to the same result. – Ben Voigt Apr 25 '14 at 23:15

Definition of derivative at $x=0$ for $f(x) = -\sqrt{x+25}$.

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$$\lim_{x\to0}\frac{5-\sqrt{x+25}}{x}=\lim_{x\to0}\frac{5-\sqrt{x+25}}{(\sqrt{x+25})²-25}=\lim_{x\to0}\frac{5-\sqrt{x+25}}{-(5-\sqrt{x+25})(5+\sqrt{x+25})}=\lim_{x\to0}\frac{-1}{5+\sqrt{x+25}}=\frac{-1}{10}$$

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Another possible method is the expand the square root term by Taylor Series (also known as Taylor expansion) after taking 25 out of the sqrt. Have you covered Taylor expansion? It's a life saver in many situations!!

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