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Can someone please tell me how to solve this, please!

lim      tan(sqrt( x))
x->0   ------------------
         x*sqrt(1+1/x)

I know that the answer is either -1 or 1, but HOW?

Big thank you (and rep) to anyone who can explain.
i looked and then:getmecoat:
Reply 2
Avatar for v-zero
v-zero
14
Apply l'Hopital's...
v-zero
Apply l'Hopital's...

It just makes it look worse. Have you any idea how i can do it step by step?
Reply 4
Avatar for v-zero
v-zero
14
learner_dancer
It just makes it look worse. Have you any idea how i can do it step by step?

Keep applying it until it looks better, more or less... :s-smilie:
learner_dancer
lim      tan(sqrt( x))
x->0   ------------------
         x*sqrt(1+1/x)

I know that the answer is either -1 or 1, but HOW?

Big thank you (and rep) to anyone who can explain.


It is L'Hopital's rule...

ddx[tan(x)]=sec2(x)2x \dfrac{d}{dx}[tan(\sqrt{x})] = \dfrac{sec^2(\sqrt{x})}{2\sqrt{x}}

x1+1x=xx+1 x\sqrt{1+\frac{1}{x}} = \sqrt{x}\sqrt{x+1}

hence (using product rule)

ddx[x1+1x]=ddx[xx+1]=x2x+1+x+12x=2x+2(x+1)4xx+1=2x+12xx+1 \dfrac{d}{dx}[x\sqrt{1+\frac{1}{x}}] = \dfrac{d}{dx}[ \sqrt{x}\sqrt{x+1}] = \dfrac{\sqrt{x}}{2\sqrt{x+1}} + \dfrac{\sqrt{x+1}}{2\sqrt{x}} = \dfrac{2x+2(x+1)}{4\sqrt{x}\sqrt{x+1}} = \dfrac{2x+1}{2\sqrt{x}\sqrt{x+1}}

So L'hoptials rule gives

lim (x->0) of

sec2(x)x+12x+1 \dfrac{sec^2(\sqrt{x})\sqrt{x+1}}{2x+1}

which gives +1. Where the -1 comes from at the moment escapes me - maybe someone else can shed light on this?

Disclaimer: I may have made an algebraic mistake here somewhere (lets face it, that's likely)

Second disclaimer: I felt an (almost) full solution was the best (and probably only) way to explain this in this instance.
spread_logic_not_hate

Disclaimer: I may have made an algebraic mistake here somewhere (lets face it, that's likely)


I don't think you did. I liked the change of form of x1+1x x \sqrt{1+\frac{1}{x}} I would have just differentiated it like that without simplifying.

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