Collection of Maths Problems

Limits of products

Task number: 3029

• Variant 1

  \( \displaystyle \lim_{x\to \infty} x\ln\left(1-\frac2x\right) \)

• Resolution

  Because \( \displaystyle \lim_{x\to \infty}\frac{\left(\ln\left(1-\frac2x\right)\right)'}{(x^{-1})'}= \lim_{x\to \infty}\frac{\frac1{1-\frac2x}\cdot\frac{-2}{-x^2}}{-x^{-2}}=-2 \)

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

  \( \displaystyle \lim_{x\to \infty} x\ln\left(1-\frac2x\right)=-2 \)

• Result

  The limit is \(-2\).

• Variant 2

  \( \displaystyle \lim_{x\to 0^+} x^n\ln(x) \)

• Resolution

  Because \( \displaystyle \lim_{x\to 0^+}\frac{(\ln x)'}{(x^{-n})'}= \lim_{x\to 0^+}\frac{\frac1x}{-nx^{-n-1}}= \lim_{x\to 0^+}-\frac1n x^n=0 \)

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

  \( \displaystyle \lim_{x\to 0^+} x^n\ln(x)=0 \)

• Result

  The limit is 0.

• Variant 3

  \( \displaystyle \lim_{x\to \infty} x e^{-x^2} \)

• Resolution

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

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

  \( \displaystyle \lim_{x\to \infty} x e^{-x^2}=0 \)

• Result

  The limit is \(0\).

• Variant 4

  \( \displaystyle \lim_{x\to \infty}x(e^\frac1x-1) \)

• Resolution

  Because \( \displaystyle \lim_{x\to \infty}\frac{\left(e^\frac1x-1\right)'}{(x^{-1})'}= \lim_{x\to \infty}\frac{-x^2 e^\frac1x}{-x^{-2}}= \lim_{x\to \infty}e^\frac1x=1 \)

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

  \( \displaystyle \lim_{x\to \infty}x(e^\frac1x-1)=1 \)

• Result

  The limit is \(1\).

• Variant 5

  \( \displaystyle \lim_{x\to \infty} x\arctan\frac1x \)

• Resolution

  Because \( \displaystyle \lim_{x\to \infty}\frac{\left(\arctan\frac1x\right)'}{(x^{-1})'}= \lim_{x\to \infty}\frac{-x^2 \frac1{1+\frac1{x^2}}}{-x^{-2}}=1 \)

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

  \( \displaystyle \lim_{x\to \infty} x\arctan\frac1x=1 \)

• Result

  The limit is 1.

• Variant 6

  \( \displaystyle \lim_{x\to \infty}x^2\arcsin\frac1x \)

• Resolution

  Because \( \displaystyle \lim_{x\to \infty}\frac{\left(\arcsin\frac1x\right)'}{(x^{-2})'}= \lim_{x\to \infty}\frac{-x^{-2} \frac1{\sqrt{1-\frac1{x^2}}}}{-2x^{-3}}= \infty \)

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

  \( \displaystyle \lim_{x\to \infty}x^2\arcsin\frac1x=\infty \)

• Result

  The limit is \(\infty\).

• Variant 7

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

• Resolution

  Because \( \displaystyle \lim_{x\to 1^-}\frac{\left(\ln(1-x)\right)'}{((\ln x)^{-1})'}= \lim_{x\to 1^-}\frac{-\frac1{1-x}}{\frac1{(\ln x)^2x}}= \lim_{x\to 1^-}\frac{x\ln^2 x}{1-x} \)

  and also \( \displaystyle \lim_{x\to 1^-}\frac{(\ln^2 x)'}{(1-x)'}= \lim_{x\to 1^-}\frac{2\ln x\cdot\frac1x}{-1}=0 \)

  and the denominators are non-zero in a neighborhood of \(1\), we have

  \( \displaystyle \lim_{x\to 1^-} \ln x \ln(1-x)=0 \)

• Result

  The limit is 0.

• Variant 8

  \( \displaystyle \lim_{x\to 0^+} \sqrt{x} \ln\sqrt{x} \)

• Resolution

  Because \( \displaystyle \lim_{x\to 0^+}\frac{\left(\frac{\ln x}2\right)'}{\left(x^{-\frac12}\right)'}= \lim_{x\to 0^+}\frac{\frac1{2x}}{-\frac12 x^{-\frac32}}= \lim_{x\to 0^+}-\sqrt{x}=0 \)

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

  \( \displaystyle \lim_{x\to 0^+} \sqrt{x} \ln\sqrt{x}=0 \)

• Result

  The limit is 0.