Collection of Maths Problems

Various limits

Task number: 2985

• Variant 1

  \(\displaystyle \lim_{x\to 0}\frac{\sin x}{2x} \)

• Resolution

  \[ \lim_{x\to 0}\frac{\sin x}{2x} = \frac12\lim_{x\to 0}\frac{\sin x}{x} = \frac12 \]

• Result

  \(\frac 12\)

• Variant 2

  \(\displaystyle \lim_{x\to 0}\frac{\sin(\sin (\sin x))}{\hbox{tg}(\hbox{tg}(x))} \)

• Resolution

  \[ \lim_{x\to 0}\frac{\sin(\sin (\sin x))}{\hbox{tg}(\hbox{tg}(x))} = \] \[ = \lim_{x\to 0}\frac{\sin(\sin (\sin x))}{\sin(\sin x)} \cdot \lim_{x\to 0} \frac{\sin (\sin x)}{\sin x} \cdot \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0} \frac{\hbox{tg} \ x}{\hbox{tg}(\hbox{tg}(x))} \cdot \lim_{x\to 0} \frac{x}{\hbox{tg} \ x} = \] \[ = 1 {\cdot} 1 \cdot 1 {\cdot} 1 \cdot 1 = 1. \]

  We have used the fact that \[ \lim_{x\to 0} \frac{\hbox{tg} \ x}{x} = 1, \] which we can easily justify as follows: \[ \lim_{x\to 0} \frac{\hbox{tg} \ x}{x} = \lim_{x\to 0} \frac{\sin x}{(\cos x)x} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0} \frac{1}{\cos x} = 1 {\cdot} 1. \] This also gives us \[ \lim_{x\to 0} \frac{x}{\hbox{tg} \ x}. \] Finally let us note that we have used the theorem about the limit of a composite function several times, and that in every case the inner function never had the value 0 in a punctured neighborhood around 0.

• Variant 3

  \(\displaystyle \lim_{x\to 0}\frac{\hbox{tg}(x) - \sin(x)}{x^3} \)

• Resolution

  \[ \lim_{x\to 0}\frac{\hbox{tg}(x) - \sin(x)}{x^3} = \lim_{x\to 0}\frac{\sin x}{x} \cdot \lim_{x\to 0} \frac{\frac{1}{\cos x} - 1}{x^2} = 1 \cdot \lim_{x\to 0} \frac{1 - \cos x}{(\cos x)x^2} = \] \[ = \lim_{x\to 0} \frac1{\cos x}\lim_{x\to 0} \frac{1 - \cos^2 x}{(1 + \cos x)x^2} = 1 \cdot \lim_{x\to 0} \frac1{1 + \cos x} \lim_{x\to 0} \frac{\sin^2 x}{x^2} = 1 \cdot \frac12 {\cdot} 1 \cdot 1 = \frac12. \]

• Variant 4

  \(\displaystyle \lim_{x\to 0}\frac{1-\cos^3(x)}{x \sin (\pi x)} \)

• Hint

  Expand \(1 - \cos^3 x\) jako \((1-\cos x)(1 + \cos x + \cos^2 x)\).

• Result

  \(\frac{3}{2\pi}\)

• Variant 5

  \(\displaystyle \lim_{x\to \infty}x\sin{\left( \frac1x\right)} \)

• Hint

  Use the theorem about the limit of a composite function with the inner function \(\frac 1x\).

• Result

  1

• Variant 6

  \(\displaystyle \lim_{x\to 0}\frac{a^x - 1}{x}, a > 0 \)

• Resolution

  \[ \lim_{x\to 0}\frac{a^x - 1}{x} = \lim_{x\to 0}\frac{\left(e^{\ln a}\right)^x - 1}{x} = \lim_{x\to 0}\frac{e^{x\ln a} - 1}{x\cdot \ln a} \cdot \ln a = 1 \cdot \ln a = \ln a. \]

• Result

  \(\ln a\)

• Variant 7

  \(\displaystyle \lim_{x\to e}\frac{\ln x - 1}{x - e} \)

• Resolution

  \[ \lim_{x\to e}\frac{\ln x - 1}{x - e} = \lim_{x\to e}\frac{\ln x - \ln e}{x - e} = \lim_{x\to e}\frac{\ln \left( \frac xe\right)}{x - e} = \lim_{x\to e}\frac{\ln \left( \frac xe\right)}{e(\frac xe - 1)} = \frac1e. \] In the steps above we used the known limit \[ \lim_{x\to 1}\frac{\ln x}{x - 1} = 1, \] keeping in mind the conditions for the limit of a composite function (namely note that \(\frac{x}e\) never has the value \(1\) in any punctured neighborhood around \(e\))!

• Variant 8

  \(\displaystyle \lim_{x\to 0^+}\frac{e^{\sqrt{\sin x}} - \cos x}{\sqrt x} \)

• Resolution

  \[ \lim_{x\to 0^+}\frac{e^{\sqrt{\sin x}} - \cos x}{\sqrt x} = \lim_{x\to 0^+}\frac{e^{2\sqrt{\sin x}} - \cos^2 x}{\underbrace{(e^{\sqrt{\sin x}} + \cos x)}_{\to 2}\sqrt x} = \frac12 \lim_{x\to 0^+}\frac{e^{2\sqrt{\sin x}} - 1 + \sin^2 x}{\sqrt x} = \] \[ = \lim_{x\to 0^+}\frac{e^{2\sqrt{\sin x}} - 1}{2\sqrt x} + \frac12 \lim_{x\to 0^+}\frac{\sin^2 x}{\sqrt x} = \lim_{x\to 0^+}\frac{e^{2\sqrt{\sin x}} - 1}{2\sqrt{\sin x}}\cdot \frac{\sqrt{\sin x}}{\sqrt x} + \frac12 \lim_{x\to 0^+}\frac{\sin^2 x}{\sqrt x} = \] \[ = 1 {\cdot} 1 + \lim_{x\to 0^+}\frac{\sqrt{\sin x}}{\sqrt x} \cdot \sin^{3/2} x = 1 + 1 {\cdot} 0 = 1. \]

• Result

  1

• Variant 9

  \(\displaystyle \lim_{x\to \infty} \sqrt{x + \sqrt{x + \sqrt x}} - \sqrt x \)

• Hint

  Use the relationship \(A - B = \frac{A^2 - B^2}{A+B}\).

• Result

  \(\frac12\)

• Variant 10

  \(\displaystyle \lim_{x\to 0} \frac{\cos\left(\frac{\sin x}x\right)} {1 + \cos\left(\frac{\sin^2 x}{x^2}\right)}. \)

• Result

  \[ \frac{\cos 1}{1 + \cos 1} \]

• Variant 11

  \(\displaystyle \lim_{x \to 0} (x + 1)^{\frac1x}. \)

• Hint

  Transform an expression of the form \(a^b\) to \(e^{a \ln b}\), or transform the expression to the limit \(\lim_{x \to \infty} \left( 1 + \frac1x\right)^x = e\).