Let g(t)=t−9t−3g\left(t\right)=\frac{t-9}{\sqrt{t}-3}g(t)=t−3t−9.Make a conjecture about the value of lim⁡t→9t−9t−3{\displaystyle\lim_{t\to9}\frac{t-9}{\sqrt{t}-3}}t→9limt−3t−9.

Ch. 2 - Limits

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

All textbooks

Briggs 3rd Edition

Ch. 2 - Limits

Problem 9b

Chapter 2, Problem 9b

Let $g$\left$(t$\right$)=$\frac{t-9}{\sqrt{t}$-3}$ .
Make a conjecture about the value of ${$\displaystyle\]\lim$_{t$\to$9}$\frac{t-9}{\sqrt{t}$-3}}$ .

Verified step by step guidance

First, observe that directly substituting t = 9 into the function g(t) = $\frac{t-9}{\sqrt{t}$-3} results in an indeterminate form $\frac{0}{0}$. This suggests that we need to simplify the expression to evaluate the limit.

To simplify, consider multiplying the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{t}$ + 3. This technique helps eliminate the square root in the denominator.

After multiplying, the expression becomes $\frac{(t-9)(\sqrt{t}$+3)}{($\sqrt{t}$-3)($\sqrt{t}$+3)}. The denominator simplifies to t - 9, as it is a difference of squares.

Now, the expression simplifies to $\frac{(t-9)(\sqrt{t}$+3)}{t-9}. Notice that the (t-9) terms in the numerator and denominator can be canceled out, provided t $\neq$ 9.

After canceling, the expression simplifies to $\sqrt{t}$ + 3. Now, substitute t = 9 into this simplified expression to find the limit, which is no longer indeterminate.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limits

A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. It helps in understanding the function's behavior near points where it may not be explicitly defined, such as points of discontinuity or indeterminate forms. In this question, we are interested in evaluating the limit of a function as t approaches 9.

Indeterminate Forms

Indeterminate forms occur in calculus when evaluating limits leads to expressions like 0/0 or ∞/∞, which do not provide clear information about the limit's value. In the given function, substituting t = 9 results in the form 0/0, necessitating further analysis, such as algebraic manipulation or L'Hôpital's Rule, to resolve the limit.

L'Hôpital's Rule

L'Hôpital's Rule is a method used to evaluate limits of indeterminate forms by differentiating the numerator and denominator separately. If the limit of a function results in an indeterminate form, applying this rule can simplify the evaluation process, allowing for a clearer determination of the limit's value. This technique is particularly useful in the context of the limit presented in the question.

Recommended video:

Guided course

5:50

Power Rules

Related Practice

Textbook Question

Assume lim x→1 f(x)=8,lim x→1 g(x)=3, and lim x→1 h(x)=2 Compute the following limits and state the limit laws used to justify your computations.

lim x→1 f(x) / g(x)−h(x)

views

Textbook Question

Determine the following limits.

lim x→1 √5x+6

Textbook Question

Assume lim x→1 f(x)=8,lim x→1 g(x)=3, and lim x→1 h(x)=2 Compute the following limits and state the limit laws used to justify your computations.

lim x→1 (f(x)−g(x))

Textbook Question

Determine the following limits.

lim x→1000 18π^2

Textbook Question

Let $g$\left$(t$\right$)=$\frac{t-9}{\sqrt{t}$-3}$ .

Make two tables, one showing values of $g$ for $t=8.9,8.99$ , and $8.999$ and one showing values of $g$ for $t=9.1,9.01$ , and $9.001$ .

Textbook Question

Suppose the rental cost for a snowboard is \$25 for the first day (or any part of the first day) plus \$15 for each additional day (or any part of a day).

Evaluate lim t→2.9 f(t).

Let g(t)=t−9t−3g\(\left\)(t\(\right\))=\(\frac{t-9}{\sqrt{t}\)-3}g(t)=t−3t−9.Make a conjecture about the value of limt→9t−9t−3{\(\displaystyle\]\lim\)_{t\(\to\)9}\(\frac{t-9}{\sqrt{t}\)-3}}t→9limt−3t−9.

Verified video answer for a similar problem:

Key Concepts

Limits

Indeterminate Forms

L'Hôpital's Rule

Let $g\(\left\)(t\(\right\))=\(\frac{t-9}{\sqrt{t}\)-3}$ .
Make a conjecture about the value of ${\(\displaystyle\]\lim\)_{t\(\to\)9}\(\frac{t-9}{\sqrt{t}\)-3}}$ .