Question

Let k(x) be piecewise function such that k(x) = sinx/x if x ≠ 0, 0 if x=0. let h(x) = 1+x from domain (-infinity, 2), and also let h(x) = -1+x from domain [1, infinity)

what would be limit as x approaches 0 of k(x)-h(x)/k(x) ?

a 0
b 1
c 2

Answer 1

a. 0

Explanation:

You want the limit of (k(x) -h(x))/k(x) as x approaches 0 when k(x) = sin(x)/x {x≠0} and h(x)=x+1 {x<1}.

Limit

Since we're concerned about the limit as x → 0, we don't have to be concerned with the fact that the expression is undefined at x = 0.

The function h(x) is defined as h(0) = 1, so we can just be concerned with the value of ...

lim[x→0] (k(x) -1)/k(x)

The limit of k(x) as x → 0 is 1, so this becomes ...

lim[x→0] (k(x) -1)/k(x) = (1 -1)/1 = 0

Sin(x)/x

At x=0, sin(x)/x is the indeterminate form 0/0, so its limit there can be found using L'Hôpital's rule. Differentiating numerator and denominator, we have ...

lim[x→0] sin(x)/x = lim[x→0] cos(x)/1 = cos(0) = 1

The fact that k(0) = 0 is irrelevant with respect to this limit.

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Additional comment

We like to use a graphing calculator to validate limit values. The attachment shows the various functions involved. It also shows that as x gets arbitrarily close to 0 from either direction, the value of g(x) does likewise. This is all that is required for (0, 0) to be declared the limit. The lack of definition of g(x) at x=0 simply means the relation has a (removable) discontinuity there.

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