Collection of Maths Problems

Limits of differences

Task number: 3030

• Variant 1

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

• Resolution

  \( \displaystyle \lim_{x\to \infty}\left(\arctan x - \frac1x\right)= \lim_{x\to \infty}\frac{x\arctan x -1}x \)

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

  and the denominator is non-zero in a neighborhood of \(\infty\) we have \( \displaystyle \lim_{x\to \infty}\left(\arctan x - \frac1x\right)=\frac{\pi}2 \)

• Result

  The limit is \(\frac{\pi}2\).

• Variant 2

  \( \displaystyle \lim_{x\to 0} \left(\frac1{3x}-\frac1{\sin x}\right) \)

• Resolution

  If the limit exists, \( \displaystyle \lim_{x\to 0} \left(\frac1{3x}-\frac1{\sin x}\right)= \lim_{x\to 0} \frac{\sin x- 3x}{3x \sin x} \)

  The denominator is non-zero in a neighborhood of \(0\), so by l'Hospital's rule

  \( \displaystyle \lim_{x\to 0} \frac{(\sin x- 3x)'}{(3x \sin x)'}= \lim_{x\to 0} \frac{\cos x- 3}{3\sin x+3x\cos x} \)

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

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

  so we have \( \displaystyle \lim_{x\to 0^+} \left(\frac1{3x}-\frac1{\sin x}\right)=-\infty \lim_{x\to 0^-} \left(\frac1{3x}-\frac1{\sin x}\right)=\infty \)

• Result

  The limit does not exist.

• Variant 3

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

• Resolution

  \( \displaystyle \lim_{x\to \infty} (x-\ln(x+1))= \lim_{x\to \infty} (\ln e^x-\ln(x+1))= \lim_{x\to \infty} \ln \left(\frac{e^x}{x+1}\right) \)

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

  and the denominator is non-zero in a neighborhood of \(\infty\) we have \( \displaystyle \lim_{x\to \infty} (x-\ln(x+1))=\infty \)

• Result

  The limit is \(\infty\).

• Variant 4

  \( \displaystyle \lim_{x\to 1} \left(\frac{m}{1-x^m}-\frac{n}{1-x^n}\right) \) for \(m,n\in \mathbb N\)

• Resolution

  \( \displaystyle \lim_{x\to 1} \left(\frac{m}{1-x^m}-\frac{n}{1-x^n}\right)= \lim_{x\to 1} \frac{m-mx^n-n-nx^m}{1-x^m-x^n+x^{m+n}} \)

  Because \( \displaystyle \lim_{x\to 1} \frac{(m-mx^n-n-nx^m)'}{(1-x^m-x^n+x^{m+n})'}= \lim_{x\to 1} \frac{mn(x^{m-1}-x^{n-1})}{(m+n)x^{m+n-1}-mx^{m-1}-nx^{n-1}} \)

  and also \( \displaystyle \lim_{x\to 1} \frac{(x^{m-1}-x^{n-1})'}{((m+n)x^{m+n-1}-mx^{m-1}-nx^{n-1})'}=\\ \lim_{x\to 1} \frac{(m-1)x^{m-2}-(n-1)x^{n-2}}{(m+n)(m+n-1)x^{m+n-2}-m(m-1)x^{m-2}-n(n-1)x^{n-2}}=\frac{m-n}{2mn} \)

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

  \( \displaystyle \lim_{x\to 1} \left(\frac{m}{1-x^m}-\frac{n}{1-x^n}\right)=mn\frac{m-n}{2mn}=\frac{m-n}2 \)

• Result

  The limit equals \(\frac{m-n}2\).