If f(x) = √5x+1, then evaluate limit h>0 {f(x+h)-f(x) ÷ h}
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f(x)=x
2
−5x+1
f(2)=2
2
−5×2+1
=4−10+1
=−6+1
=−5
f(−1)=(−1)
2
−5×−1+1
=1+5+1
=7
f(
3
1
)=
3
1
2
−5×
3
1
+1
=
9
1
−
3
5
+1
=
9
1
−
9
15
+
9
9
=
9
10−15
=
9
−5
f(2)−f(−1)+f(
3
1
)=−5−7−
9
5
=−12−
9
5
=
9
−108−5
=−
9
113
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