\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Limits Involving Inverse Trig Functions

Limits

Using Limits


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To find the limits of inverse trig functions, you need to be very familiar with how inverse trig functions work. Remember that \(\sin^{-1}(x) \neq 1/\sin(x)\) in this context. Go to the precalculus section on inverse trig functions if you need some review.

There are several ways to evaluate limits of inverse trig functions. Of course the most direct way is substitution, which you want to always try first. If that doesn't work, you can move the limit inside the inverse trig function like this.

\(\displaystyle{ \lim_{ x\to c }{ \arcsin(x) } = \arcsin \left[ \lim_{ x\to c }{ x } \right] }\)

This works with the other inverse functions as well since the inverse trig functions are continuous on their restricted domains. (Not sure what we mean? Review this page.)

Another idea is to convert to regular trig functions and try to see what will happen. So, for example, \(\theta = \arcsin(x) \to \sin(\theta) = x\). This will help since you are probably more familar with straight trig functions than you are with inverse trig functions. You may also be able to use trig identities to simplify the equation so that you can then evaluate the limit.

A third idea that will help you see what is going on is to graph the inverse trig function. You don't need to know the graphs off the top of your head since you probably already know the graphs of the trig functions. Just graph the trig functions and flip the graph about the line \(y=x\). In practice problem 2311, this is what he does. Although this technique is probably not enough work to get full credit for finding the limit, it can really help you visualize what is going on and may give you an idea on how to solve the problem.

Okay, try these practice problems.

Practice

Unless otherwise instructed, evaluate these limits. Give your answers in exact terms.

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

Problem Statement

Evaluate this limit. Give your answer in exact terms. \(\displaystyle{ \lim_{x \to -1^+}{\sin^{-1}(x)} }\)

Solution

2311 video

video by PatrickJMT

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\(\displaystyle{ \lim_{x \to \infty}{\arcsin\left[ \frac{1+x^2}{1+2x^2} \right] } }\)

Problem Statement

Evaluate this limit. Give your answer in exact terms. \(\displaystyle{ \lim_{x \to \infty}{\arcsin\left[ \frac{1+x^2}{1+2x^2} \right] } }\)

Solution

2312 video

video by PatrickJMT

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\(\displaystyle{ \lim_{x \to \infty}{\arctan(e^{3x})} }\)

Problem Statement

Evaluate this limit. Give your answer in exact terms. \(\displaystyle{ \lim_{x \to \infty}{\arctan(e^{3x})} }\)

Solution

2313 video

video by PatrickJMT

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

Problem Statement

Evaluate this limit. Give your answer in exact terms. \(\displaystyle{ \lim_{x \to 0^+}{ \arctan(\ln(x))} }\)

Solution

2314 video

video by PatrickJMT

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inverse trig limits 17calculus youtube playlist

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, evaluate these limits. Give your answers in exact terms.

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