High School Math : Using Limits with Continuity

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Calculus Ii — Integrals

Function

 

The above graph depicts a function \(\displaystyle h(x)\). Does \(\displaystyle \lim_{x\rightarrow 1} h(x)\) exist, and why or why not?

Possible Answers:

\(\displaystyle \lim_{x\rightarrow 1} h(x)\) exists because \(\displaystyle \lim_{x\rightarrow 1^{-}} | h(x)| = \lim_{x\rightarrow 1^{+}} |h(x)|\)

\(\displaystyle \lim_{x\rightarrow 1} h(x)\) does not exist because \(\displaystyle \lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)\).

\(\displaystyle \lim_{x\rightarrow 1} h(x)\) does not exist because \(\displaystyle f (1 ) \neq \lim_{x\rightarrow 1^{-}} h(x)\)

\(\displaystyle \lim_{x\rightarrow 1} h(x)\) exists because \(\displaystyle f (1 ) = \frac{\lim_{x\rightarrow 1^{-}} h(x) + \lim_{x\rightarrow 1^{+ }} h(x)}{2}\)

\(\displaystyle \lim_{x\rightarrow 1} h(x)\) does not exist because \(\displaystyle f (1 ) \neq \lim_{x\rightarrow 1^{+}} h(x)\)

Correct answer:

\(\displaystyle \lim_{x\rightarrow 1} h(x)\) does not exist because \(\displaystyle \lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)\).

Explanation:

\(\displaystyle \lim_{x\rightarrow 1} h(x)\) exists if and only if \(\displaystyle \lim_{x\rightarrow 1^{-}} h(x) = \lim_{x\rightarrow 1^{+}} h(x)\). As can be seen from the diagram, \(\displaystyle \lim_{x\rightarrow 1^{-}} h(x) = 1\), but \(\displaystyle \lim_{x\rightarrow 1^{+}} h(x) = -1\). Since \(\displaystyle \lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)\),   \(\displaystyle \lim_{x\rightarrow 1} h(x)\) does not exist.

Example Question #1 : Limits

Function

The above graph depicts a function \(\displaystyle f(x)\). Does \(\displaystyle \lim_{x\rightarrow 1} f(x)\) exist, and why or why not?

Possible Answers:

\(\displaystyle \lim_{x\rightarrow 1} f(x)\) exists because \(\displaystyle \lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)\)

\(\displaystyle \lim_{x\rightarrow 1} f(x)\) does not exist because \(\displaystyle f\) is not continuaous at \(\displaystyle x = 1\).

\(\displaystyle \lim_{x\rightarrow 1} f(x)\) does not exist because \(\displaystyle \lim_{x\rightarrow 1^{+}} f(x) \neq \lim_{x\rightarrow 1^{-}} f(x)\)

\(\displaystyle \lim_{x\rightarrow 1} f(x)\) does not exist because \(\displaystyle f(1) \neq \lim_{x\rightarrow 1^{-}} f(x)\)

\(\displaystyle \lim_{x\rightarrow 1} f(x)\) does not exist because \(\displaystyle f(1) \neq \lim_{x\rightarrow 1^{+}} f(x)\)

Correct answer:

\(\displaystyle \lim_{x\rightarrow 1} f(x)\) exists because \(\displaystyle \lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)\)

Explanation:

\(\displaystyle \lim_{x\rightarrow 1} f(x)\) exists if and only if \(\displaystyle \lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)\);

the actual value of \(\displaystyle f(1)\) is irrelevant, as is whether \(\displaystyle f\) is continuous there.

As can be seen,

\(\displaystyle \lim_{x\rightarrow 1^{+}} f(x) = 1\) and \(\displaystyle \lim_{x\rightarrow 1^{-}} f(x) = 1\);

therefore, \(\displaystyle \lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)\),

and \(\displaystyle \lim_{x\rightarrow 1} f(x)\) exists.

Example Question #1 : Calculus Ii — Integrals

A function is defined by the following piecewise equation:

\(\displaystyle f(x)=\left\{\begin{matrix} x^{2}+4x-3, x< 3\\10x-12, x\geq 3 \end{matrix}\right.\)

At \(\displaystyle x=3\), the function is:

Possible Answers:

discontinuous

continuous

Correct answer:

continuous

Explanation:

The first step to determine continuity at a point is to determine if the function is defined at that point. When we substitute in 3 for \(\displaystyle x\), we get 18 as our \(\displaystyle y\)-value. \(\displaystyle x=3\) is thus defined for this function.

The next step is determine if the limit of the function is defined at that point. This means that the left-hand limit must be equal to the right-hand limit at \(\displaystyle x=3\). Substitution reveals the following:

\(\displaystyle \lim_{x\rightarrow 3^{-}}f(x)=(3)^{2}+4(3)-3=9+12-3=18\)

\(\displaystyle \lim_{x\rightarrow 3^{+}}f(x)=10(3)-12=18\)

Both sides of the function, therefore, approach a \(\displaystyle y\)-value of 18.

 

Finally, we must ensure that the curve is smooth by checking the limit of the derivative of both sides.

\(\displaystyle \lim_{x\rightarrow 3^{-}}f^{'}(x)=2(3)+4=10\)

\(\displaystyle \lim_{x\rightarrow 3^{+}}f^{'}(x)=10\)

 

Since the function passes all three tests, it is continuous.

Example Question #1 : Limits

Function

The graph depicts a function \(\displaystyle g(x)\). Does \(\displaystyle \lim_{x\rightarrow 1} g(x)\) exist?

Possible Answers:

\(\displaystyle \lim_{x\rightarrow 1} g(x)\) exists because \(\displaystyle g(x)\) is constant on \(\displaystyle (-\infty , 1)\).

\(\displaystyle \lim_{x\rightarrow 1} g(x)\) does not exist because \(\displaystyle g(1)\) is undefined.

\(\displaystyle \lim_{x\rightarrow 1} g(x)\) does not exist because \(\displaystyle g\) is not continuous at \(\displaystyle x = 1\).

\(\displaystyle \lim_{x\rightarrow 1} g(x)\) exists because \(\displaystyle \lim_{x\rightarrow 1^{-}} g(x) = \lim_{x\rightarrow 1^{+}} g(x)\).

\(\displaystyle \lim_{x\rightarrow 1} g(x)\) does not exist because \(\displaystyle \lim_{x\rightarrow 1^{-}} g(x) \neq \lim_{x\rightarrow 1^{+}} g(x)\).

Correct answer:

\(\displaystyle \lim_{x\rightarrow 1} g(x)\) exists because \(\displaystyle \lim_{x\rightarrow 1^{-}} g(x) = \lim_{x\rightarrow 1^{+}} g(x)\).

Explanation:

\(\displaystyle \lim_{x\rightarrow 1}g(x)\) exists if and only if \(\displaystyle \lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)\); the actual value of \(\displaystyle f(1)\) is irrelevant.

As can be seen, \(\displaystyle \lim_{x\rightarrow 1^{+}} g(x) = 1\) and \(\displaystyle \lim_{x\rightarrow 1^{-}}g(x) = 1\); therefore, \(\displaystyle \lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)\), and \(\displaystyle \lim_{x\rightarrow 1} g(x)\)  exists.

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