Different ways of computing a limit (Part 5)

Calculus, Geometry 1 Minute

One of my colleagues placed the following problem on an exam for his Calculus II course…

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

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #5. Another geometric approach. The numbers x and \sqrt{x^2+1} can be viewed as two sides of a right triangle with legs 1 and x and hypotenuse \sqrt{x^2+1}. Therefore, the length of the hypotenuse must be larger than the length of one leg but less than the sum of the lengths of the two legs. In other words,

x < \sqrt{x^2+1} < x+1,

or

1 < \displaystyle \frac{\sqrt{x^2+1}}{x} < \displaystyle 1+\frac{1}{x}.

 Clearly \displaystyle \lim_{x \to \infty} 1 = 1 and \displaystyle \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right) = 1. Therefore, by the Sandwich Theorem, we can conclude that \displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = 1.

Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

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