How do you use the limit definition of the derivative to find the derivative of #f(x)=x/(x+1)#?

1 Answer
Jim H
Sep 18, 2016

See below for two possible solutions.

Explanation:

Soultion 1

Using: #f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h#

#f'(x) = lim_(hrarr0)([(x+h)/(x+h+1)]-[x/(x+1)])/h#

If we try substitution, we get the indeterminate form #0/0#. Let's get a single fraction on top over #h/1#, then multiply the top by the reciprocal of the bottom (invert and multiply).

# = lim_(hrarr0)(((x+h)(x+1)-x(x+h+1))/((x+h+1)(x+1))) / (h/1)#

# = lim_(hrarr0)(((x^2+x+xh+h-x^2-xh-x))/((x+h+1)(x+1))) / (h/1)#

# = lim_(hrarr0)((h)/((x+h+1)(x+1))) / (h/1)#

# = lim_(hrarr0)((h)/(x+h+1)(x+1)) * 1/h#

# = lim_(hrarr0) 1/((x+h+1)(x+1)) #.

Now, when we evaluate, we do not get #0# in the denominator, so the limit is

# = 1/(x+1)^2 #.

#f'(x) = 1/(x+1)^2 #.

Solution 2

Using #f'(x) = lim_(trarrx)(f(t)-f(x))/(t-x)#

#lim_(trarrx)(f(t)-f(x))/(t-x) = lim_(trarrx)([t/(t+1)]-[x/(x+1)])/(t-x)#

If we try substitution, we get the indeterminate form #0/0#. Let's get a single fraction on top over #(t-x)/1#, then multiply the top by the reciprocal of the bottom (invert and multiply).

# = lim_(trarrx)((t(x+1)-x(t+1))/((t+1)(x+1))) / ((t-x)/1)#

# = lim_(trarrx)(((tx+t-tx-x))/((t+1)(x+1))) / ((t-x)/1)#

# = lim_(trarrx)((t-x)/((t+1)(x+1))) / ((t-x)/1)#

# = lim_(trarrx)(t-x)/((t+1)(x+1)) * 1/(t-x)#

# = lim_(trarrx) 1/((t+1)(x+1)) #.

Now, when we evaluate, we do not get #0# in the denominator, so the limit is

# = 1/(x+1)^2 #.

#f'(x) = 1/(x+1)^2 #.

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