How to solve limits with substitution

 
 
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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.

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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???.

 
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