How do you use the limit definition to find the slope of the tangent line to the graph #f(x) = sqrt(x+1)# at (8,3)?

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Answer 1 · 2016-06-21T12:51:36

see below: #1/6#

#f'(x) = (sqrt{x + h + 1} - sqrt{x + 1})/ h ]_{h\to 0}#

# = ( (sqrt{x + h + 1} + sqrt{x + 1}) (sqrt{x + h + 1} - sqrt{x + 1} ) )/ ((sqrt{x + h + 1} + sqrt{x + 1}) h) ]_{h\to 0} #

# = ( x + h + 1 - (x + 1)) / ((sqrt{x + h + 1} + sqrt{x + 1}) h) ]_{h\to 0} #

# = ( h) / ((sqrt{x + h + 1} + sqrt{x + 1}) h) ]_{h\to 0} #

# = ( 1) / ((sqrt{x + h + 1} + sqrt{x + 1}) ) ]_{h\to 0} #

now #sqrt{x + h + 1) ]_{h\to 0} = sqrt{x + 1)#

So the limit is:
# ( 1) / (2sqrt{x + 1} ) #

# ( 1) / (2sqrt{8 + 1} ) = 1/6#