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# Limits by direct substitution

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)

## Video transcript

so let's see if we can find the limit as X approaches negative 1 of 6x squared plus 5x minus 1 now the first thing that might jump out at you is this right over here this expression could be used to define the graph of a parabola and when you think about this I'm not doing a rigorous proof here parabola would look something like this this would be an upward-opening parabola look something like this it is a this graph visually is continuous you don't see any jumps or gaps in it and in general upper out of a quadratic like this it's going to be defined for all values of X for all real numbers it's going to be continuous for all real numbers and so something is continuous for all real numbers well then the limit as X approaches some real number is going to be the same thing as just evaluating the expression at that real number so what am i saying I'm just going to say it another way we know that some function is continuous is continuous at some x value at x equals a if and only if I'll write that as if or if if and only if the limit as X approaches a of f of X is equal to F of is equal to f of a so I didn't do a rigorous proof here but just it's conceptually not a big jump to say ok well this is this is just a standard quadratic right over here it's defined for all real numbers and in fact it's continuous for all real numbers and so we know that this expression it could define a continuous function so that means that the limit as X approaches a for this expression is just the same thing as evaluating this expression at a in this case our a is negative 1 so all I have to do is evaluate this at negative 1 this is going to be 6 times negative 1 squared plus 5 times negative 1 minus 1 so that's just 1 this is negative 5 so it's 6 minus 5 minus 1 which is equal to 0 and we are done
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