How to solve limits with substitution
Substitution is the simplest way to evaluate a limit, but it doesn’t always work
As we’ve seen in previous lessons, the simplest way to evaluate a limit is to substitute the value we’re approaching into the function.
For instance, given the function ???f(x)=x+1???, finding the limit as ???x\to5??? is as easy as substituting ???x=5??? into ???f(x)???.
???\lim_{x\to5}(x+1)???
???5+1???
???6???
If ???f(x)??? is an expression that contains only polynomials, roots, absolute values, exponentials, logarithms, trig or inverse trig functions, then we may be able to evaluate using substitution, and we’ll have
???\lim_{x\to a}f(x)=f(a)???
But if the function is undefined at ???x=a???, or if ???x=a??? is the transition point between two pieces of a piecewise-defined function, then we can’t apply the substitution rule.
Nevertheless, when we evaluate a limit we should always try substitution first before any other technique, because it’s the easiest and fastest method. If substitution doesn’t work, then we can try evaluating the limit by a different method.
How to solve limit problems using substitution
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Substituting into a polynomial function
Example
Evaluate the limit.
???\lim_{x\to -2}(x^2+2x+6)???
Since we’re approaching ???x=-2???, we’ll substitute ???x=-2??? into the function.
???(-2)^2+2(-2)+6???
???4-4+6???
???6???
So the limit of the function as ???x=-2??? is ???6???.