How do you find the limit of #2x^2+4x+1# as #x->-3#?

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Answer 1 · 2017-12-11T15:42:18

#7#

The limit can be found by just plugging in the value #x=-3#

#2(-3)^2+4(-3)+1=7#

#lim_(x->-3)(2x^2+4x+1)=7#

Answer 2 · 2017-12-11T15:54:43

#7#

#"we can evaluate the limit by direct substitution"#

#rArrlim_(xto-3)2x^2+4x+1#

#=2(-3)^2+4(-3)+1=7#