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Precalculus : Find the Limit of a Function

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Example Questions

Example Question #16 : Limits

The Michaelis-Menten equation is important in chemical kinetics. Suppose we are given the following equation relating K (reaction rate) and C (concentration):

$K(C) = \tfrac{1000C}{500 + C}$ $\displaystyle K(C) = \tfrac{1000C}{500 + C}$

Determine:

$\lim_{C \rightarrow \infty } K(C)$ $\displaystyle \lim_{C \rightarrow \infty } K(C)$

Possible Answers:

1000 $\displaystyle 1000$

$\displaystyle 2$

Limit Does Not Exist

$\infty$ $\displaystyle \infty$

500 $\displaystyle 500$

Correct answer:

1000 $\displaystyle 1000$

Explanation:

There are a number of ways to solve this. Either we can solve by finding what K(C) evaluates to for larger values of C and see where they converge, i.e. :

K(10000) = 952.38

K(1000000) = 999.5

etc...

And we see that K(C) approaches 1000 for larger concentrations, C.

Or we can notice that the dominant term in the numerator is 1000C; dominant term in the denominator is C.

1000C / C = 1000, which will ultimately be our limit.

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Example Question #11 : Find The Limit Of A Function

Let $f(x)=\frac{x+7}{x^{2}-2}$ $\displaystyle f(x)=\frac{x+7}{x^{2}-2}$.

Find $\lim_{x\rightarrow 2}$ $\displaystyle \lim_{x\rightarrow 2}$ f(x) $\displaystyle f(x)$.

Possible Answers:

$-\frac{3}{4}$ $\displaystyle -\frac{3}{4}$

$\frac{9}2{}$ $\displaystyle \frac{9}2{}$

$\displaystyle 1$

$\displaystyle 0$

Undefined

Correct answer:

$\frac{9}2{}$ $\displaystyle \frac{9}2{}$

Explanation:

To find $\lim_{x\rightarrow 2}$ $\displaystyle \lim_{x\rightarrow 2}$ f(x) $\displaystyle f(x)$ here, you need only plug in $\displaystyle 2$ for $\displaystyle x$:

$f(x)=\frac{x+7}{x^{2}-2}=\frac{2+7}{2^{2}-2}=\frac{9}{2}$ $\displaystyle f(x)=\frac{x+7}{x^{2}-2}=\frac{2+7}{2^{2}-2}=\frac{9}{2}$

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Example Question #11 : Find The Limit Of A Function

Evaluate

$\lim_{-3}\frac{\frac{1}{x}+\frac{1}{3}}{1+\frac{2}{x+1}}$ $\displaystyle \lim_{-3}\frac{\frac{1}{x}+\frac{1}{3}}{1+\frac{2}{x+1}}$.

Possible Answers:

$-\frac{2}{5}$ $\displaystyle -\frac{2}{5}$

$\frac{3}{5}$ $\displaystyle \frac{3}{5}$

$\frac{4}{9}$ $\displaystyle \frac{4}{9}$

$\frac{2}{9}$ $\displaystyle \frac{2}{9}$

$-\frac{1}{3}$ $\displaystyle -\frac{1}{3}$

Correct answer:

$\frac{2}{9}$ $\displaystyle \frac{2}{9}$

Explanation:

Find a common denominator for both the upper and lower expressions and then simplify:

$\\ \lim_{x\rightarrow -3}\frac{\frac{1}{x}+\frac{1}{3}}{1+\frac{2}{x+1}}\\ \\=\lim_{x\rightarrow -3}\frac{\frac{3+x}{3x}}{\frac{x+1+2}{x+1}}\\ \\=\lim_{x\rightarrow -3}\frac{3+x}{3x}\times\frac{x+1}{3+x}\\ \\=\lim_{x\rightarrow -3}\frac{x+1}{3x}\\ \\=\frac{-3+1}{3(-3)}\\ \\=\frac{-2}{-9}\\ \\=\frac{2}{9}$ $\displaystyle \\ \lim_{x\rightarrow -3}\frac{\frac{1}{x}+\frac{1}{3}}{1+\frac{2}{x+1}}\\ \\=\lim_{x\rightarrow -3}\frac{\frac{3+x}{3x}}{\frac{x+1+2}{x+1}}\\ \\=\lim_{x\rightarrow -3}\frac{3+x}{3x}\times\frac{x+1}{3+x}\\ \\=\lim_{x\rightarrow -3}\frac{x+1}{3x}\\ \\=\frac{-3+1}{3(-3)}\\ \\=\frac{-2}{-9}\\ \\=\frac{2}{9}$

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Example Question #11 : Find The Limit Of A Function

Find the limit of the function:

$\lim_{x\rightarrow 5^{+}} \frac{5}{5-x}$ $\displaystyle \lim_{x\rightarrow 5^{+}} \frac{5}{5-x}$

Possible Answers:

The limit can't be determined.

$\frac{1}{5}$ $\displaystyle \frac{1}{5}$

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

$\displaystyle 5$

Correct answer:

$-\infty$ $\displaystyle -\infty$

Explanation:

Substituting the value of x=5 $\displaystyle x=5$ will yield $\frac{5}{0}$ $\displaystyle \frac{5}{0}$, which is not in indeterminate form. Therefore, L'Hopital's rule cannot be used.

The question asks for the limit as x approaches to five from the right side in the graph. As the graph approaches closer and closer to x=5 $\displaystyle x=5$, the y-value decreases to negative infinity and will never touch x=5 $\displaystyle x=5$.

The correct answer is: $-\infty$ $\displaystyle -\infty$

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Example Question #11 : Find The Limit Of A Function

Find the limit.

$\lim_{x\rightarrow 3}\frac{2x-6}{x-3}$ $\displaystyle \lim_{x\rightarrow 3}\frac{2x-6}{x-3}$

Possible Answers:

$\displaystyle =3$

Does not exist

$\displaystyle =0$

$\displaystyle =2$

=12 $\displaystyle =12$

Correct answer:

$\displaystyle =2$

Explanation:

In the unsimplified form, the limit does not exist; however, the numerator can be factored and simplified.

$\lim_{x\rightarrow 3}\frac{2x-6}{x-3}$ $\displaystyle \lim_{x\rightarrow 3}\frac{2x-6}{x-3}$

$=\lim_{x\rightarrow 3}\frac{2(x-3)}{x-3}$ $\displaystyle =\lim_{x\rightarrow 3}\frac{2(x-3)}{x-3}$

$\displaystyle =2$

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Example Question #11 : Find The Limit Of A Function

Find the limit: $\lim_{x \to 6} \frac{x-3}{3}$ $\displaystyle \lim_{x \to 6} \frac{x-3}{3}$

Possible Answers:

$\displaystyle 3$

$\displaystyle 2$

$\displaystyle 1$

$\displaystyle 6$

$\textup{Undefined}$ $\displaystyle \textup{Undefined}$

Correct answer:

$\displaystyle 1$

Explanation:

The first and only step is to substitute the value of $\displaystyle x$ into the function. Since there is not a zero denominator or an indeterminate form, we do not have to worry about L'Hopital or an undefined limit.

$\lim_{x \to 6} \frac{x-3}{3} = \frac{6-3}{3} = \frac{3}{3} = 1$ $\displaystyle \lim_{x \to 6} \frac{x-3}{3} = \frac{6-3}{3} = \frac{3}{3} = 1$

The limit will approach to $\displaystyle 1$.

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Example Question #12 : Find The Limit Of A Function

Find the following limit:

$\lim_{x\rightarrow \infty} \frac{3x^2+1}{2x^2}$ $\displaystyle \lim_{x\rightarrow \infty} \frac{3x^2+1}{2x^2}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$\displaystyle undefined$

$\displaystyle 0$

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

Correct answer:

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

Explanation:

To solve, simply realize you are dealing with a limit whose numerator and demominator have the same max power. Thus the limit is simply the division of their coefficient.

$\frac{3x^2+1}{2x^2}\approx \frac{3x^2}{2x^2}=\frac{3}{2}$ $\displaystyle \frac{3x^2+1}{2x^2}\approx \frac{3x^2}{2x^2}=\frac{3}{2}$

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Precalculus : Find the Limit of a Function

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #16 : Limits

Example Question #11 : Find The Limit Of A Function

Example Question #11 : Find The Limit Of A Function

Example Question #11 : Find The Limit Of A Function

Example Question #11 : Find The Limit Of A Function

Example Question #11 : Find The Limit Of A Function

Example Question #12 : Find The Limit Of A Function

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