Collection of Maths Problems

Limit of a composite function

Task number: 2983

• Variant 1

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

• Resolution

  Let \(\displaystyle g(x)=1+\frac{1}{x^2}\), \(f(y)=\sqrt{y} \).

  We can easily determine that \(\displaystyle \lim_{x\to\infty}g(x)=\lim_{x\to\infty}(1+\frac{1}{x^2})=1 \) and \(\displaystyle \lim_{y\to 1}f(y)=\lim_{y\to 1}\sqrt{y}=1. \)

  Also, the function \(f\) is continuous at the point \(y=1\), and so by the theorem about the limit of a composite function we obtain \(\displaystyle \lim_{x\to \infty}\sqrt{1+\frac{1}{x^2}}=\lim_{x\to \infty}f(g(x))=1 \).

• Variant 2

  \(\displaystyle \lim_{x\to 0}\frac{\log(1+\sin x)}{x} \)

• Hint

  Expand the fraction with the expression \(\frac{\sin x}{x}\),

• Resolution

  Part of the given expression reminds us of the well-known limit \(\displaystyle \lim_{y\to 0}\frac{\log(1+y)}{y}=1, \) and so we transform the given function in the punctured neighborhood around \(0\) to \(\displaystyle \frac{\log(1+\sin x)}{x}=\frac{\log(1+\sin x)}{\sin x}\frac{\sin x}{x}. \)

  So the given limit is equal to

  \(\displaystyle \lim_{x\to 0}\frac{\log(1+\sin x)}{x} =\lim_{x\to 0}\frac{\log(1+\sin x)}{\sin x}\lim_{x\to 0}\frac{\sin x}{x} =\lim_{x\to 0}\frac{\log(1+\sin x)}{\sin x}. \)

  Now we use the theorem about the limit of a composite function for \(g(x)=\sin x\), \(f(y)=\frac{\log(1+y)}{y}\).

  We know that \(\displaystyle \lim_{x\to 0}g(x)=0\), and that \(\displaystyle \lim_{y\to 0}f(y)=1\).

  What is more, in the punctured neighborhood \(P(0{,}1)\) we have \(g(x)=\sin x\neq 0\). So by the theorem about the limit of a composite function, the limit on the right side equals \(1\), and so the given limit equals \(1\).