Question

Evaluate the given limit. $$ \lim _{t \rightarrow 5}\left\langle 2 t+1,3 t^{2}-1, \sin t\right\rangle $$

Short answer

The limit is \( \langle 11, 74, \sin(5) \rangle \).

Step-by-step solution

Identify the Components of the Vector Function

The function is a vector function: \( \langle 2t + 1, 3t^2 - 1, \sin t \rangle \). It is composed of three separate component functions: \( 2t + 1 \), \( 3t^2 - 1 \), and \( \sin t \). We need to evaluate the limit of each component separately as \( t \rightarrow 5 \).

Compute the Limit of the First Component

The first component is \( 2t + 1 \). To find its limit as \( t \rightarrow 5 \), substitute 5 into the expression: \[ 2(5) + 1 = 10 + 1 = 11 \] Thus, the limit of the first component is 11.

Compute the Limit of the Second Component

The second component is \( 3t^2 - 1 \). Calculate its limit as \( t \rightarrow 5 \) by substituting 5: \[ 3(5)^2 - 1 = 3(25) - 1 = 75 - 1 = 74 \] Therefore, the limit of the second component is 74.

Compute the Limit of the Third Component

The third component is \( \sin t \). Evaluate its limit as \( t \rightarrow 5 \) by substituting 5: \[ \sin(5) \] Since \( \sin t \) is continuous, the limit is \( \sin(5) \). This is a constant, approximately \( -0.95892 \), if evaluated numerically.

Combine the Limits of All Components

Combine the limits of each component function to find the overall limit of the vector function: \[ \lim _{t \rightarrow 5} \left\langle 2t + 1, 3t^2 - 1, \sin t \right\rangle = \left\langle 11, 74, \sin(5) \right\rangle \]Thus, the limit of the vector function is \( \langle 11, 74, \sin(5) \rangle \).

Key concepts

Vector Functions

Vector functions are mathematical tools that associate each input with a vector, rather than a single scalar. They are written using angle brackets, such as \( \langle f(t), g(t), h(t) \rangle \), where each component, like \( f(t) \), is a function dependent on the variable \( t \).
These components represent the vector in multi-dimensional space, giving a more dynamic representation of data compared to single-variable functions. For example, the vector function \( \langle 2t + 1, 3t^2 - 1, \sin t \rangle \) defines a vector path as \( t \) varies.
The importance of vector functions grows in fields involving movement and change across multiple directions, such as physics and engineering. They give us a powerful means to describe physical trajectories, multi-dimensional fields, and various other phenomena.

• Each component must be evaluated separately to understand the change of the overall vector.
• Each component function handles transformations along one axis in vector space.

Thus, understanding the individual components aids in interpreting the behavior and properties of the entire vector function.

Limits of Functions

The concept of limits is fundamental in calculus and vital for analyzing the behavior of functions as they approach a certain point. For vector functions, evaluating limits involves assessing each component separately.
In the example \( \langle 2t+1, 3t^2-1, \sin t \rangle \), we find the limit by substituting \( t = 5 \) into each function. This means:

• For \( 2t+1 \), as \( t \rightarrow 5 \), the expression becomes \( 11 \).


• For \( 3t^2-1 \), substituting gives \( 74 \).


• For \( \sin t \), we get \( \sin(5) \) which is approximately \( -0.95892 \).

The evaluation of these limits shows how each individual component behaves as \( t \) moves closer to the chosen value. Once these separate evaluations are completed, we can combine these to get the overall limit of the vector function.
This process is crucial as it enables us to predict and understand system behavior at specific instances, simplifying analyses in practical applications.

Continuity in Calculus

Continuity is a key property in calculus, indicating that a function smoothly progresses without any sudden jumps or breaks. A function is continuous if it is defined at a point and if the limit as you approach that point is equal to the function's value at that point.
For our vector function \( \langle 2t+1, 3t^2-1, \sin t \rangle \), each component is continuous, because they're made up of continuous functions: linear, quadratic, and sinusoidal functions.

• The linear function \( 2t+1 \) is continuous at all real numbers.


• The quadratic function \( 3t^2-1 \) is also continuous everywhere.


• The sine function, \( \sin t \), is known for its continuous wave-like pattern without breaks.

Overall, if each component of a vector function is continuous, the entire vector function is continuous as well. This assures smooth transitions and helps us understand the flow and progression through vector spaces without interruption.
Continuity ensures predictable behavior, making notation and calculations in calculus simpler and more reliable.