Collection of Maths Problems

Limits of ratios

Task number: 3028

• Variant 1

  \( \displaystyle \lim_{x\to \infty}\frac{\ln x}{x} \)

• Resolution

  Because \( \displaystyle \lim_{x\to \infty}\frac{(\ln x)'}{x'}= \lim_{x\to \infty}\frac{\frac1x}{1}= \lim_{x\to \infty}\frac{1}{x}=0 \)

  and the denominator is non-zero in a neighborhood of \(\infty\), we have

  \( \displaystyle \lim_{x\to \infty}\frac{\ln x}{x}=0 \)

• Result

  The limit is 0.

• Variant 2

  \( \displaystyle \lim_{x\to 0}\frac{\ln(\cos x)}{x} \)

• Resolution

  Because \( \displaystyle \lim_{x\to 0}\frac{(\ln(\cos x))'}{x'}= \lim_{x\to 0}\frac{\frac{-\sin x}{\cos x}}{1}=0 \)

  and the denominator is non-zero in a neighborhood of \(0\), we have

  \( \displaystyle \lim_{x\to 0}\frac{\ln(\cos x)}{x}=0 \)

• Result

  The limit is 0.

• Variant 3

  \( \displaystyle \lim_{x\to \infty}\frac{\ln(1+x^2)}{\ln(2+3x^3)} \)

• Resolution

  Because \( \displaystyle \lim_{x\to \infty}\frac{(\ln(1+x^2))'}{(\ln(2+3x^3))'}= \lim_{x\to \infty}\frac{\frac{2x}{4x^2}}{\frac{9x^2}{2+3x^2}}= \lim_{x\to \infty}\frac{2(2+3x^2)}{9x(1+x^2)}=\frac 23 \)

  and the denominator is non-zero in a neighborhood of \(\infty\), we have

  \( \displaystyle \lim_{x\to \infty}\frac{\ln(1+x^2)}{\ln(2+3x^3)}=\frac 23 \)

• Result

  The limit is \(\frac 23\).

• Variant 4

  \( \displaystyle \lim_{x\to 0}\frac{e^{x^2}-1}{\cos x-1} \)

• Resolution

  Because \( \displaystyle \lim_{x\to 0}\frac{(e^{x^2}-1)'}{(\cos x-1)'}= \lim_{x\to 0}\frac{2xe^{x^2}}{-\sin x}=-2 \)

  and the denominator is non-zero in a neighborhood of \(0\), we have

  \( \displaystyle \lim_{x\to 0}\frac{e^{x^2}-1}{\cos x-1}=-2 \)

• Result

  The limit is \(-2\).

• Variant 5

  \( \displaystyle \lim_{x\to 0^+}\frac{\ln(\sin2 x)}{\ln(\sin x)} \)

• Resolution

  Because \( \displaystyle \lim_{x\to 0^+}\frac{(\ln(\sin2 x))'}{(\ln(\sin x))'}= \lim_{x\to 0^+}\frac{2\frac{\cos 2x}{\sin2 x}}{\frac{\cos x}{\sin x}}= 2\lim_{x\to 0^+}\frac{\sin x}{\sin2 x} \)

  and also \( \displaystyle \lim_{x\to 0^+}\frac{\cos x}{2\cos 2x}= \lim_{x\to 0^+}\frac{(\sin x)'}{(\sin2 x)'}=\frac12 \)

  and the denominators are non-zero in a neighborhood of \(0\), we have

  \( \displaystyle \lim_{x\to 0^+}\frac{\ln(\sin2 x)}{\ln(\sin x)}=2\cdot\frac12=1 \)

• Result

  The limit is 1.

• Variant 6

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

• Resolution

  Because \( \displaystyle \lim_{x\to 0}\frac{(\sin x)'}{(\arcsin x)'}= \lim_{x\to 0}\cos x \cdot \sqrt{1-x^2}=1 \)

  and the denominator is non-zero in a neighborhood of \(0\), we have

  \( \displaystyle \lim_{x\to 0}\frac{\sin x}{\arcsin x}=1 \)

• Result

  The limit is 1.

• Variant 7

  \( \displaystyle \lim_{x\to 0}\frac{e^x-1}{\sin 2x} \)

• Resolution

  Because \( \displaystyle \lim_{x\to 0}\frac{(e^x-1)'}{(\sin 2x)'}= \lim_{x\to =}\frac{e^x}{2\cos 2x}=\frac12 \)

  and the denominator is non-zero in a neighborhood of \(0\), we have

  \( \displaystyle \lim_{x\to 0}\frac{e^x-1}{\sin 2x}=\frac12 \)

• Result

  The limit is \(\frac12\).

• Variant 8

  \( \displaystyle \lim_{x\to \infty}\frac{e^{\sqrt{x}}}{x} \)

• Resolution

  Because \( \displaystyle \lim_{x\to \infty}\frac{(e^{\sqrt{x}})'}{x'}= \lim_{x\to \infty}\frac{e^{\sqrt{x}}\frac1{2\sqrt{x}}}{1}= 2\lim_{x\to \infty}\frac{e^{\sqrt{x}}}{\sqrt{x}} \)

  and also \( \displaystyle \lim_{x\to \infty}\frac{(e^{\sqrt{x}})'}{(\sqrt{x})'}= \lim_{x\to \infty}\frac{e^{\sqrt{x}}\frac1{2\sqrt{x}}}{\frac1{2\sqrt{x}}}= 2\lim_{x\to \infty}{e^{\sqrt{x}}}=\infty \)

  and the denominators are non-zero in a neighborhood of \(\infty\), we have

  \( \displaystyle \lim_{x\to \infty}\frac{e^{\sqrt{x}}}{x}=\infty \)

• Result

  The limit is \(\infty\).

• Variant 9

  \( \displaystyle \lim_{x\to 0}\frac{\tan x}{\cos x-1} \)

• Resolution

  Because the denominator is non-zero in a neighborhood of \(\infty\) and

  \( \displaystyle \lim_{x\to 0}\frac{(\tan x)'}{(\cos x-1)'}= \lim_{x\to 0}\frac{\frac 1{\cos^2 x}}{\sin x} \)

  where of course \( \displaystyle \lim_{x\to 0^+}\frac{1}{\sin x}=\infty,\ \lim_{x\to 0^-}\frac{1}{\sin x}=-\infty \)

  we obtain \( \displaystyle \lim_{x\to 0^+}\frac{\tan x}{\cos x-1}=\infty,\ \lim_{x\to 0^-}\frac{\tan x}{\cos x-1}=-\infty \)

• Result

  The limit does not exist.

• Variant 10

  \( \displaystyle \lim_{x\to \infty}\frac{\sin x-x}{x^3} \)

• Resolution

  Because \( \displaystyle \lim_{x\to \infty}\frac{(\sin x-x)'}{(x^3)'}= \lim_{x\to \infty}\frac{\cos x-1}{3x^2}=0 \)

  and the denominator is non-zero in a neighborhood of \(\infty\), we have

  \( \displaystyle \lim_{x\to \infty}\frac{\sin x-x}{x^3}=0 \)

• Result

  The limit is 0.

• Variant 11

  \( \displaystyle \lim_{x\to 0}\frac{\arctan(1+x)-\arctan(1-x)}{x} \)

• Resolution

  Because \( \displaystyle \lim_{x\to 0}\frac{(\arctan(1+x)-\arctan(1-x))'}{x'}= \lim_{x\to 0}\frac{\frac1{1+(1+x)^2}-\frac1{1+(1-x)^2}}{1}=0 \)

  and the denominator is non-zero in a neighborhood of \(0\), we have

  \( \displaystyle \lim_{x\to 0}\frac{\arctan(1+x)-\arctan(1-x)}{x}=0 \)

• Variant 12

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