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How to calculate following with out using L'Hospital rule

$$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$

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\begin{eqnarray*} \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sqrt{\pi}-\sqrt{\arccos x}}{\sqrt{x+1}} & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\pi-\arccos x}{2\sqrt{\pi}\sqrt{x+1}}\\ & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sin\left(\pi-\arccos x\right)}{2\sqrt{\pi}\sqrt{x+1}}\\ & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sin\arccos x}{2\sqrt{\pi}\sqrt{x+1}}\\ & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sqrt{1-x^{2}}}{2\sqrt{\pi}\sqrt{x+1}}\\ & = & \frac{1}{\sqrt{2\pi}} \end{eqnarray*}

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Let $\sqrt{\arccos(x)} = t$. We then have $x = \cos(t^2)$. Since $x \to (-1)^+$, we have $t^2 \to \pi^-$. Hence, we have $$\lim_{x \to (-1)^+} \dfrac{\sqrt{\pi} - \sqrt{\arccos(x)}}{\sqrt{1+x}} = \overbrace{\lim_{t \to \sqrt{\pi}^-} \dfrac{\sqrt{\pi} - t}{\sqrt{1+\cos(t^2)}}}^{t = \sqrt{\arccos(x)}} = \underbrace{\lim_{y \to 0^+} \dfrac{y}{\sqrt{1+\cos((\sqrt{\pi}-y)^2)}}}_{y = \sqrt{\pi}-t}$$ $$1+\cos((\sqrt{\pi}-y)^2) = 1+\cos(\pi -2 \sqrt{\pi}y + y^2) = 1-\cos(2 \sqrt{\pi}y - y^2) = 2 \sin^2 \left(\sqrt{\pi} y - \dfrac{y^2}2\right)$$ Hence, \begin{align} \lim_{y \to 0^+} \dfrac{y}{\sqrt{1+\cos((\sqrt{\pi}-y)^2)}} & = \dfrac1{\sqrt2} \lim_{y \to 0^+} \dfrac{y}{\sin \left(\sqrt{\pi}y - \dfrac{y^2}2\right)}\\ & = \dfrac1{\sqrt2} \lim_{y \to 0^+} \dfrac{\left(\sqrt{\pi}y - \dfrac{y^2}2\right)}{\sin \left(\sqrt{\pi}y - \dfrac{y^2}2\right)} \dfrac{y}{\left(\sqrt{\pi}y - \dfrac{y^2}2\right)} = \dfrac1{\sqrt{2 \pi}} \end{align}

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Provided the following limits exist, \begin{align*} A &= \lim_{x \to -1^{+}}\left(\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{\sqrt{x+1}} \right)\\ &= \lim_{x\to -1^+}\left(\frac{\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{x + 1}}{\frac{\sqrt{x+1} - 0}{x + 1}} \right)\\ &= \frac{\lim_{x\to -1^+}\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{x + 1}}{\lim_{x\to -1^+}\frac{\sqrt{x+1} - 0}{x + 1}}, \end{align*} which we recognize as the quotient of the "derivative from the right" of $\sqrt{\cos^{-1} x}$ and $\sqrt{ x + 1}$ at $x = -1$. So, \begin{align*} A &= \lim_{x\to -1^+}\frac{\frac{1}{2\sqrt{1 - x^2}\sqrt{\cos^{-1} x}}}{\frac{1}{2\sqrt{x + 1}}}\\ &= \lim_{x\to -1^+}\frac{\sqrt{x + 1}}{\sqrt{1 - x^2}\sqrt{\cos^{-1} x}}\\ &= \lim_{x\to -1^+}\frac{\sqrt{x + 1}}{\sqrt{1 - x}\sqrt{1 + x}\sqrt{\cos^{-1} x}}\\ &= \lim_{x\to -1^+}\frac{1}{\sqrt{1 - x}\sqrt{\cos^{-1} x}}\\ &= \frac{1}{\sqrt{2\pi}}.\\ \end{align*} (in the first equation we flip the sign of the derivative because they're being taken in different orders: the top has the value first, limit second, bottom has limit first, value second.) Admittedly, I'm playing pretty fast and loose here, but if you're ambitious and have a bit of tenacity (and free time), you can probably fill in the missing details.

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Hint

$$ \lim_{x \to (-1)^+} \cos^{-1} x = \lim_{ h \to 0 } {\cos^{-1} (-1 + h) }$$

From here, you can follow this previous question.

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