Q: Let f:R\to R be a differentiable function satisfying f'(3)-f'(2)=0. Then \displaystyle\lim_{x\to 0}\dfrac{[1+f(3+x)-f(3)]^{1/x}}{[1+f(2+x)-f(2)]^{1/x}} is equal to

1. As x\to0, \dfrac{f(3+x)-f(3)}{x}\to f'(3), so the numerator \to e^{f'(3)}; similarly the denominator \to e^{f'(2)}. The limit is e^{f'(3)-f'(2)}=e^0=1.

Q: \displaystyle\lim_{x\to 1^-}\dfrac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}} is equal to

\sqrt{\dfrac{2}{\pi}}. Put x=\sin\theta; as x\to1^-,\ \theta\to\pi/2^-. With t=\pi/2-\theta\to0^+, \sqrt{\pi}-\sqrt{2\sin^{-1}x}=\sqrt\pi-\sqrt{\pi-2t}\approx t/\sqrt\pi and \sqrt{1-x}=\sqrt{1-\cos t}\approx t/\sqrt2, so the ratio \to\sqrt{2/\pi}.

Q: Let [x] denote the greatest integer less than or equal to x. Then \displaystyle\lim_{x\to 0}\dfrac{\tan(\pi\sin^2x)-(|x|-\sin(x[x]))^2}{x^2}

does not exist. For x\to0^+, [x]=0, so |x|-\sin(x[x])=x and the limit becomes \pi-1. For x\to0^-, [x]=-1, so |x|-\sin(x[x])=-x-\sin(-x)=-x+\sin x, which is O(x^3), making that piece vanish and the limit becomes \pi. Since the two one-sided limits disagree, the limit does not exist.

Q: For each t\in R, let [t] be the greatest integer less than or equal to t. Then \displaystyle\lim_{x\to 1^-}\dfrac{(1-|x|+\sin|1-x|)\sin\!\left(\dfrac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}

equals 0. As x\to1^-, 0<1-x<1 so [1-x]=0 throughout, which makes \sin\!\left(\frac{\pi}{2}[1-x]\right)=\sin 0=0 identically for all such x, forcing the whole expression to be 0 near x=1^-; hence the limit is 0.

Q: For each x\in R, let [x] be the greatest integer less than or equal to x. Then \displaystyle\lim_{x\to 0^-}\dfrac{x([x]-|x|)\sin[x]}{|x|} is equal to

-\sin 1. For x\to0^-, [x]=-1 and |x|=-x, so [x]-|x|\to-1 and \sin[x]=\sin(-1)=-\sin1. The expression equals \dfrac{x(-1)(-\sin1)}{-x}=-\sin1.

Q: For each t\in R, let [t] be the greatest integer less than or equal to t. Then \displaystyle\lim_{x\to 0^+} x\left(\left[\dfrac1x\right]+\left[\dfrac2x\right]+\cdots+\left[\dfrac{15}{x}\right]\right)

is equal to 120. Since \dfrac{k}{x}-1 0 and summing k=1 to 15 squeezes the expression to 1+2+\cdots+15=120.

Q: Let f(x)=\left[1+\dfrac{x(1-|1-x|)}{|1-x|}\right]\cos\!\left(\dfrac{1}{1-x}\right) for x\ne1. Then

\displaystyle\lim_{x\to1^-}f(x) does not exist. For x\to1^-, |1-x|=1-x, so the bracket simplifies to a value bounded away from 0, while \cos\!\left(\dfrac{1}{1-x}\right) oscillates between -1 and 1 infinitely often as x\to1^-; hence the limit does not exist.

Q: Let p=\displaystyle\lim_{x\to0^+}(1+\tan^2\sqrt{x})^{1/2x}, then \log p is equal to

\dfrac12. \tan^2\sqrt x\sim x as x\to0^+, so (1+\tan^2\sqrt x)^{1/2x}\to e^{\lim \tan^2\sqrt x/(2x)}=e^{1/2}. Thus p=e^{1/2} and \log p=\dfrac12.

Q: Let \alpha(a) and \beta(a) be the roots of the equation (\sqrt[3]{1+a}-1)x^2+(\sqrt{1+a}-1)x+(\sqrt[6]{1+a}-1)=0, where a>-1. Then \displaystyle\lim_{a\to0^+}\alpha(a) and \displaystyle\lim_{a\to0^+}\beta(a) are

-\dfrac12 and -1. As a\to0^+, \sqrt[n]{1+a}-1\sim a/n, so the coefficients scale in the ratio \frac13:\frac12:\frac16=2:3:1, giving the limiting equation 2x^2+3x+1=0, whose roots are -1 and -\dfrac12.

Question 1

Let f:R	o R be a differentiable function satisfying f'(3)-f'(2)=0. Then \displaystyle\lim_{x	o 0}\dfrac{[1+f(3+x)-f(3)]^{1/x}}{[1+f(2+x)-f(2)]^{1/x}} is equal to

Accepted Answer

1. As x	o0, \dfrac{f(3+x)-f(3)}{x}	o f'(3), so the numerator 	o e^{f'(3)}; similarly the denominator 	o e^{f'(2)}. The limit is e^{f'(3)-f'(2)}=e^0=1.

Question 2

\displaystyle\lim_{x	o 1^-}\dfrac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}} is equal to

Accepted Answer

\sqrt{\dfrac{2}{\pi}}. Put x=\sin	heta; as x	o1^-,\ 	heta	o\pi/2^-. With t=\pi/2-	heta	o0^+, \sqrt{\pi}-\sqrt{2\sin^{-1}x}=\sqrt\pi-\sqrt{\pi-2t}\approx t/\sqrt\pi and \sqrt{1-x}=\sqrt{1-\cos t}\approx t/\sqrt2, so the ratio 	o\sqrt{2/\pi}.

Question 3

Let [x] denote the greatest integer less than or equal to x. Then \displaystyle\lim_{x	o 0}\dfrac{	an(\pi\sin^2x)-(|x|-\sin(x[x]))^2}{x^2}

Accepted Answer

does not exist. For x	o0^+, [x]=0, so |x|-\sin(x[x])=x and the limit becomes \pi-1. For x	o0^-, [x]=-1, so |x|-\sin(x[x])=-x-\sin(-x)=-x+\sin x, which is O(x^3), making that piece vanish and the limit becomes \pi. Since the two one-sided limits disagree, the limit does not exist.

Question 4

For each t\in R, let [t] be the greatest integer less than or equal to t. Then \displaystyle\lim_{x	o 1^-}\dfrac{(1-|x|+\sin|1-x|)\sin\!\left(\dfrac{\pi}{2}[1-x]ight)}{|1-x|[1-x]}

Accepted Answer

equals 0. As x	o1^-, 0<1-x<1 so [1-x]=0 throughout, which makes \sin\!\left(\frac{\pi}{2}[1-x]ight)=\sin 0=0 identically for all such x, forcing the whole expression to be 0 near x=1^-; hence the limit is 0.

Question 5

For each x\in R, let [x] be the greatest integer less than or equal to x. Then \displaystyle\lim_{x	o 0^-}\dfrac{x([x]-|x|)\sin[x]}{|x|} is equal to

Accepted Answer

-\sin 1. For x	o0^-, [x]=-1 and |x|=-x, so [x]-|x|	o-1 and \sin[x]=\sin(-1)=-\sin1. The expression equals \dfrac{x(-1)(-\sin1)}{-x}=-\sin1.

Question 6

For each t\in R, let [t] be the greatest integer less than or equal to t. Then \displaystyle\lim_{x	o 0^+} x\left(\left[\dfrac1xight]+\left[\dfrac2xight]+\cdots+\left[\dfrac{15}{x}ight]ight)

Accepted Answer

is equal to 120. Since \dfrac{k}{x}-1<\left[\dfrac kx\right]\le\dfrac kx, multiplying by x>0 and summing k=1 to 15 squeezes the expression to 1+2+\cdots+15=120.

Question 7

Let f(x)=\left[1+\dfrac{x(1-|1-x|)}{|1-x|}ight]\cos\!\left(\dfrac{1}{1-x}ight) for x
e1. Then

Accepted Answer

\displaystyle\lim_{x	o1^-}f(x) does not exist. For x	o1^-, |1-x|=1-x, so the bracket simplifies to a value bounded away from 0, while \cos\!\left(\dfrac{1}{1-x}ight) oscillates between -1 and 1 infinitely often as x	o1^-; hence the limit does not exist.

Question 8

Let p=\displaystyle\lim_{x	o0^+}(1+	an^2\sqrt{x})^{1/2x}, then \log p is equal to

Accepted Answer

\dfrac12. 	an^2\sqrt x\sim x as x	o0^+, so (1+	an^2\sqrt x)^{1/2x}	o e^{\lim 	an^2\sqrt x/(2x)}=e^{1/2}. Thus p=e^{1/2} and \log p=\dfrac12.

Question 9

Let \alpha(a) and \beta(a) be the roots of the equation (\sqrt[3]{1+a}-1)x^2+(\sqrt{1+a}-1)x+(\sqrt[6]{1+a}-1)=0, where a>-1. Then \displaystyle\lim_{a	o0^+}\alpha(a) and \displaystyle\lim_{a	o0^+}\beta(a) are

Accepted Answer

-\dfrac12 and -1. As a	o0^+, \sqrt[n]{1+a}-1\sim a/n, so the coefficients scale in the ratio \frac13:\frac12:\frac16=2:3:1, giving the limiting equation 2x^2+3x+1=0, whose roots are -1 and -\dfrac12.

Question 10

If \displaystyle\lim_{x	o0}[1+x\log(1+b^2)]^{1/x}=2b\sin^2	heta, b>0, and 	heta\in(-\pi,\pi], then the value of 	heta is

Accepted Answer

\pm\dfrac{\pi}{2}. The limit equals e^{\log(1+b^2)}=1+b^2. So 2b\sin^2	heta=1+b^2\ge2b by AM-GM (equality forced since \sin^2	heta\le1), which requires \sin^2	heta=1, i.e. 	heta=\pm\dfrac{\pi}{2}.

Question 11

For x>0, \displaystyle\lim_{x	o0}\left\{(\sin x)^{1/x}+\left(\dfrac1xight)^{\sin x}ight\} is

Accepted Answer

1. As x	o0^+, 0<\sin x<1 raised to the infinite power 1/x gives (\sin x)^{1/x}	o0; and \left(\dfrac1xight)^{\sin x}=e^{\sin x\ln(1/x)}	o e^0=1 since x\ln(1/x)	o0. The sum 	o1.

Question 12

Let f:R	o R be such that f(1)=3 and f'(1)=6. Then \displaystyle\lim_{x	o0}\left(\dfrac{f(1+x)}{f(1)}ight)^{1/x} equals

Accepted Answer

e^2. \left(\dfrac{f(1+x)}{f(1)}ight)^{1/x}=\left(1+\dfrac{f(1+x)-f(1)}{f(1)}ight)^{1/x}	o e^{f'(1)/f(1)}=e^{6/3}=e^2.

Question 13

For x\in R, \displaystyle\lim_{x	o\infty}\left(\dfrac{x-3}{x+2}ight)^{x} is equal to

Accepted Answer

e^{-5}. \left(\dfrac{x-3}{x+2}ight)^x=\left(1-\dfrac{5}{x+2}ight)^x	o e^{-5}.

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