Different ways of computing a limit (Part 3)

Calculus 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 #3. A trigonometric identity. When we see \sqrt{x^2+1} inside of an integral, one kneejerk reaction is to try the trigonometric substitution x = \tan \theta. So let’s use this here. Also, since x \to \infty, we can change the limit to be \theta \to \pi/2:

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\tan^2 \theta+1}}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{\sqrt{\sec^2 \theta}}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{ \sec \theta}{\tan \theta}

= \displaystyle \lim_{\theta \to \pi/2} \frac{ ~~\displaystyle \frac{1}{\cos \theta} ~~}{ ~~ \displaystyle \frac{\sin \theta}{\cos \theta} ~~ }

= \displaystyle \lim_{\theta \to \pi/2} \frac{ 1}{\sin \theta}

= 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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