What is the #lim_(h rarr 0) (arcsin(a+h) - arcsin(a))/h = 2# ?

I know I'll get an indeterminate form on substituting 0. I cross-multiplied to get the left side = 2h. I still can't find an answer...it is an MC question for which the answer is (root 3)/2.

2 Answers
Noah G
Mar 20, 2018

See below.

Explanation:

Recall that the definition of the derivative is

#f'(x) = lim_(h->0) (f(x + h) - f(x))/h#

Thus

#y= arcsinx#

Then the derivative is given as follows

#siny = x#

#cosy(dy/dx) = 1#

#dy/dx = 1/cosy#

We know that #sin^2y + cos^2y = 1#.

#dy/dx =1/sqrt(1- sin^2y)#

#dy/dx= 1/sqrt(1 -x^2)#

From the limit statement we know that #f'(a) = 2#.

#2 = 1/sqrt(1 - x^2)#

#2sqrt(1 - x^2) = 1#

#4(1 - x^2) = 1#

#4 - 4x^2 = 1#

#3 = 4x^2#

#3/4 = x^2#

#sqrt(3)/2 = x#

As required.

Hopefully this helps!

Steve M
Mar 20, 2018

In general,

# lim_(h rarr 0) (arcsin(a+h)-arcsin(a))/h = 1/(sqrt(1-a^2)) #

And if

# lim_(h rarr 0) (arcsin(a+h)-arcsin(a))/h = 2 => a = sqrt(3)/2 #

Explanation:

Let us first find a general expression for the limit:

# L = lim_(h rarr 0) (arcsin(a+h)-arcsin(a))/h #

The astute reader will spot that this limit is the same limkit used in the definition of the limit, so in fact this limit is that of

# d/(da) arcsin(a) #

ie it is the derivative of #arcsin(a)# via first principles. Let us perform two substitutions:

# {: (theta = arcsin(a+h)), (phi=arcsin(a)) :} <=> {: (sin theta = a+h), (sin phi=a) :}#

And we also note that with this substitution:

# sin theta - sin phi = a+h -a => h = sin theta - sin phi #

And so as:

# h rarr 0 => sin theta - sin phi rarr 0 => theta - phi rarr 0 #

So substituting into the limit, we can write:

# L = lim_(theta - phi rarr 0) (theta-phi)/(sin theta - sin phi) #

We can use the trigonometric identity:

# sin(A-B) -= 2cos((A+B)/2)sin((A-B)/2) #

And so we get:

# L = lim_(theta - phi rarr 0) (theta-phi)/(2cos((theta+phi)/2)sin((theta-phi)/2)) #

# \ \ = lim_(theta - phi rarr 0) 2/(2cos((theta+phi)/2)(sin((theta-phi)/2))/((theta-phi)/2)) #

# \ \ = lim_(theta - phi rarr 0) 2/(2cos((theta+phi)/2)) lim_(theta - phi rarr 0) (sin((theta-phi)/2))/((theta-phi)/2) #

We note that the second limit is a standard calculus result:

# lim_(x rarr 0) sin(x)/x = 1 #

Thus we have:

# L = lim_(theta - phi rarr 0) 2/(2cos((theta+phi)/2)) #

# \ \ = 2/(2cos((2phi)/2)) #

# \ \ = 1/(cosphi) #

# \ \ = 1/(sqrt(1-sin^2phi)) #

# \ \ = 1/(sqrt(1-a^2)) #

Giving us the standard calculus result:

# d/(da) arcsin(a) = 1/(sqrt(1-a^2)) #

We now seek the value of #a# for which

# 1/(sqrt(1-a^2)) = 2#

# => 2sqrt(1-a^2) = 1#

# :. 4(1-a^2) = 1#

# :. 1-a^2 = 1/4#

# :. a^2 = 3/4#

# :. a = sqrt(3)/2 #

Where we take the positive root to get the principal value

Related questions
Impact of this question
13421 views around the world
You can reuse this answer
Creative Commons License