Find $\lim_{x \to 0}\frac{sin{\pi \cdot \sqrt{x+1}}}{x}$

Question

Calculate: $$\lim_{x \to 0}\frac{\sin{(\pi \cdot \sqrt{x+1})}}{x}$$

I don't how to use L'Hopital so I tried to take $X=\sqrt{x+1}$ so when $x \to 0$ we have $X \to 1$.

But I can't find the real form.

Let $f(x)=\sin(\pi \sqrt{x+1})$. Then $f(0)=0$, so you're looking for $\lim_{x\to 0} \dfrac{f(x)-f(0)}{x}$. But that's just the definition of the derivative at $x=0$... — Semiclassical, Feb 28 '16 at 13:32
@user315918 If you don't know about the derivative, you should't be trying to take limits like this. Study basic differentiation first, you'll get much more out of that. — Bobson Dugnutt, Feb 28 '16 at 13:46
@Semiclassical I think you should post your elegant solution as an answer. — Therkel, Feb 28 '16 at 13:52

score 2 · Answer 1 · answered Feb 28 '16 at 13:41

Substitute $\;y:=\sqrt{x+1}\;$ , so $\;x\to 0\implies y\to 1\;$ :

$$\lim_{y\to1}\frac{\sin\pi y}{y^2-1}=\lim_{y\to1}\frac{\sin(\pi(y-1)+\pi)}{(y-1)(y+1)}=$$

$$=\lim_{y\to1}\frac{-\sin(\pi(y-1))}{\pi(y-1)}\cdot\frac\pi{y+1}=(-1)\cdot\frac\pi2=-\frac\pi2$$

score 1 · Answer 2 · answered Feb 28 '16 at 14:02

Not an answer to the stated question, but L'Hôpital's rule really would make life easier here.

In this case, $$f(x) = \sin(\pi \sqrt{x+1}), \space g(x) = x$$ so $$\frac{d f(x)}{d x} = f'(x) = \cos(\pi \sqrt{x + 1})\frac{\pi}{2\sqrt{x+1}}$$ using the chain rule for $\frac{d}{d x} f(g(x)) = f'(g(x)) g'(x)$, plus $\frac{d}{d x}\sin(x) = \cos(x)$, and $\frac{d}{d x}\sqrt{x+1} = \frac{1}{2\sqrt{x+1}}$. Finally,$$\frac{d g(x)}{d x} = g'(x) = 1$$ Since both $f'(x)$ and $g'(x)$ exist and $g'(0) \ne 0$, we only need to apply L'Hôpital's rule once: $$\lim \limits_{x \to 0} \frac{\sin(\pi \sqrt{x+1})}{x} = \lim \limits_{x \to 0} \frac{\pi \cos(\pi \sqrt{x+1})}{2 \sqrt{x + 1}}$$ which can be directly evaluated at $x=0$: $$ = \frac{ \pi \cos(\pi) }{2} = -\frac{\pi}{2} $$

score 1 · Answer 3 · answered Feb 28 '16 at 14:05

Since $\lim_{x\to0}\pi\sqrt{x+1}=\pi$, I suggest substituting $$ t=\pi(1-\sqrt{x+1}) $$ so the numerator becomes $\sin(\pi-t)=\sin t$. Next $$ x=\left(1-\frac{t}{\pi}\right)^2-1=\frac{t^2-2\pi t}{\pi^2} $$ so the limit becomes $$ \lim_{t\to0}\frac{\pi^2}{t-2\pi}\frac{\sin t}{t} $$

On the other hand, the given limit is the derivative at $0$ of the function $f(x)=\sin(\pi\sqrt{x+1})$ and the chain rule gives $$ f'(x)=\cos(\pi\sqrt{x+1})\cdot\frac{\pi}{2\sqrt{x+1}} $$ so $$ f'(0)=\cos\pi\cdot\frac{\pi}{2}=-\frac{\pi}{2} $$

score 0 · Answer 4 · answered Feb 28 '16 at 13:38

$$\lim_{x \to 0}\frac{\sin{(\pi \cdot \sqrt{x+1})}}{x}=\lim_{x \to 0}\frac{\sin{(\pi-\pi \cdot \sqrt{x+1})}}{x}=\lim_{x \to 0}\frac{\sin{(\pi(1- \sqrt{x+1}))}}{x}=\lim_{x \to 0}\frac{\sin{(\pi(1- \sqrt{x+1}))}}{(\pi(1- \sqrt{x+1}))}\frac{(\pi(1- \sqrt{x+1}))}{x}=\lim_{x \to 0}\frac{\sin{(\pi(1- \sqrt{x+1}))}}{(\pi(1- \sqrt{x+1}))}\lim_{x \to 0}\frac{(\pi(1- \sqrt{x+1}))}{x}=\lim_{x \to 0}\frac{(\pi(1- \sqrt{x+1}))}{x}=\lim_{x \to 0}\frac{(\pi(1- \sqrt{x+1}))}{x}.\frac{(1+ \sqrt{x+1})}{(1+\sqrt{x+1})}=\lim_{x \to 0}\frac{\pi(1- x-1)}{x(1+\sqrt{x+1})}=\lim_{x \to 0}\frac{\pi(- x)}{x(1+\sqrt{x+1})}=-\frac\pi2$$

score 0 · Answer 5 · answered Feb 28 '16 at 14:54

Let $\sqrt{x+1}=y+1\implies x=y^2+2y$

$$\lim_{x \to 0}\frac{\sin{(\pi \cdot \sqrt{x+1})}}{x}$$

$$=\lim_{y\to0}\dfrac{\sin\pi(y+1)}{\pi y}\cdot\lim_{y\to0}\dfrac\pi{y+2}$$

$$=-\lim_{y\to0}\dfrac{\sin\pi y}{\pi y}\cdot\dfrac\pi{0+2}=?$$ as $\sin(\pi+A)=-\sin A$

Find $\lim_{x \to 0}\frac{sin{\pi \cdot \sqrt{x+1}}}{x}$

5 Answers5

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