Problem 54
Question
Evaluate the following limits using a table of values. $$\begin{aligned}&\text { For } g(x)=\left\\{\begin{array}{ll}3 \tan \left[\frac{\pi}{4}(x+2)\right] & x \leq-1 \\\\\sqrt{x^{2}+8} & x>-1\end{array},\text { find }\right.\\\&\text { a. } \lim _{x \rightarrow-1^{-}} g(x) \quad \text { b. } \lim _{x \rightarrow-1}+g(x)\end{aligned}$$
Step-by-Step Solution
Verified Answer
Both limits are 3, so \( \lim_{x \rightarrow -1} g(x) = 3 \).
1Step 1: Evaluate the left-hand limit
To find \( \lim_{x \rightarrow -1^{-}} g(x) \), evaluate \( g(x) \) for values of \( x \) approaching \(-1\) from the left. Since \( x \leq -1 \) uses the \( 3 \tan\left[\frac{\pi}{4}(x+2)\right] \) formula, calculate \( g(x) \) for \( x = -1.5, -1.2, -1.1 \), etc.\For example, at \( x = -1.5 \):\\[g(-1.5) = 3 \tan\left[\frac{\pi}{4}(-1.5+2)\right] = 3 \tan\left[\frac{\pi}{4}(0.5)\right] = 3 \tan\left[\frac{\pi}{8}\right].\\]Evaluate similar calculations for other such points and observe the trend as \( x \) approaches \(-1\) from the left.
2Step 2: Calculate values for left-hand limit
Now, compute the actual values for each of the points identified in Step 1:- For \( x = -1.5 \), \( g(x) \approx 3 \times 0.414 = 1.242 \).- For \( x = -1.2 \), \( g(x) \approx 3 \times 0.568 = 1.704 \).- For \( x = -1.1 \), \( g(x) \approx 3 \times 0.749 = 2.247 \).These results suggest that as \( x \rightarrow -1^{-} \), \( g(x) \rightarrow 3 \).
3Step 3: Conclude left-hand limit
Based on the calculations in the previous steps, the limit from the left is \( \lim_{x \rightarrow -1^{-}} g(x) = 3 \).
4Step 4: Evaluate the right-hand limit
To find \( \lim_{x \rightarrow -1^{+}} g(x) \), evaluate \( g(x) \) for values of \( x \) approaching \(-1\) from the right using \( g(x) = \sqrt{x^2 + 8} \). Calculate \( g(x) \) for \( x = -0.9, -0.8, -0.7 \), etc.\For example, at \( x = -0.9 \):\\[g(-0.9) = \sqrt{(-0.9)^2 + 8} = \sqrt{0.81 + 8} = \sqrt{8.81} \approx 2.97.\\]Continue similarly for the other points.
5Step 5: Calculate values for right-hand limit
Calculate values for points as \( x \) approaches \(-1 \) from the right:- For \( x = -0.9 \), \( g(x) \approx 2.97 \).- For \( x = -0.8 \), \( g(x) \approx 2.92 \).- For \( x = -0.7 \), \( g(x) \approx 2.86 \).These results show that as \( x \rightarrow -1^{+} \), \( g(x) \rightarrow 3 \).
6Step 6: Conclude right-hand limit
Based on Step 5's calculations, the limit from the right is \( \lim_{x \rightarrow -1^{+}} g(x) = 3 \).
7Step 7: Conclude the limits
The left-hand and right-hand limits are both \( 3 \), thus \( \lim_{x \rightarrow -1} g(x) \) exists and equals \( 3 \).
Key Concepts
Left-Hand LimitRight-Hand LimitPiecewise Function
Left-Hand Limit
The left-hand limit of a function involves evaluating what value the function approaches as the input, or variable, approaches a specified point from the left side. In mathematical notation, we express this as \( \lim_{x \rightarrow c^{-}} f(x) \), where \( c \) is the point of approach. Consider the piecewise function defined for different ranges of \( x \), which means the way you determine the limit depends on which segment or piece of the function you are examining.
In the original problem, we have a piecewise function \( g(x) \) with specific formulas under different conditions. For \( x \leq -1 \), the formula \( 3 \tan\left[ \frac{\pi}{4}(x+2) \right] \) is used. To find the left-hand limit as \( x \) approaches \(-1\), we use values less than \(-1\) (such as \(-1.5, -1.2, -1.1\)).
Computing the function with these approaching values shows a trend or convergence towards a particular value. Through these calculations, it becomes evident that \( \lim_{x \rightarrow -1^{-}} g(x) = 3 \).
This indicates from the left direction, the function values are getting closer and closer to \( 3 \). This process is crucial in analyzing the behavior of piecewise functions near points of interest.
In the original problem, we have a piecewise function \( g(x) \) with specific formulas under different conditions. For \( x \leq -1 \), the formula \( 3 \tan\left[ \frac{\pi}{4}(x+2) \right] \) is used. To find the left-hand limit as \( x \) approaches \(-1\), we use values less than \(-1\) (such as \(-1.5, -1.2, -1.1\)).
Computing the function with these approaching values shows a trend or convergence towards a particular value. Through these calculations, it becomes evident that \( \lim_{x \rightarrow -1^{-}} g(x) = 3 \).
This indicates from the left direction, the function values are getting closer and closer to \( 3 \). This process is crucial in analyzing the behavior of piecewise functions near points of interest.
Right-Hand Limit
The right-hand limit refers to evaluating the potential value approached by a function as \( x \) gets closer to a certain point from the right. This is denoted by \( \lim_{x \rightarrow c^{+}} f(x) \). Just like with left-hand limits, examining right-hand limits involves using the appropriate piece of a piecewise function for the range of \( x \) slightly greater than \( c \).
In our example, for \( x > -1 \), the function is \( \sqrt{x^2 + 8} \). To analyze \( \lim_{x \rightarrow -1^{+}} g(x) \), we use values like \(-0.9, -0.8, -0.7\) which are just greater than \(-1\). By substituting these values into the function, we find that the values approach \( 3 \) from the right.
The calculation shows that as \( x \rightarrow -1^{+} \), \( g(x) \rightarrow 3 \), demonstrating that the function approaches \( 3 \) from the right direction.
Understanding the right-hand limit is essential for comprehending how a function behaves as it nears a particular point from greater values of \( x \), especially in functions with distinct parts.
In our example, for \( x > -1 \), the function is \( \sqrt{x^2 + 8} \). To analyze \( \lim_{x \rightarrow -1^{+}} g(x) \), we use values like \(-0.9, -0.8, -0.7\) which are just greater than \(-1\). By substituting these values into the function, we find that the values approach \( 3 \) from the right.
The calculation shows that as \( x \rightarrow -1^{+} \), \( g(x) \rightarrow 3 \), demonstrating that the function approaches \( 3 \) from the right direction.
Understanding the right-hand limit is essential for comprehending how a function behaves as it nears a particular point from greater values of \( x \), especially in functions with distinct parts.
Piecewise Function
A piecewise function can be recognized by its structure, which involves different expressions or formulas applicable over distinct intervals of \( x \). Essentially, the function's definition divides the domain into separate pieces, where each specifies a different formula depending on the range of \( x \).
In the given exercise, \( g(x) \) is defined differently depending on whether \( x \leq -1 \) or \( x > -1 \). Such a setup indicates the use of a piecewise function with two branches:
Each section of a piecewise function must be evaluated separately, and limits need to be considered from both the left and right, as demonstrated in finding \( \lim_{x \rightarrow -1} g(x) \).
Recognizing the structure and evaluating each section individually while considering the appropriate interval helps in fully understanding such functions as exemplified by these kinds of mathematical problems.
In the given exercise, \( g(x) \) is defined differently depending on whether \( x \leq -1 \) or \( x > -1 \). Such a setup indicates the use of a piecewise function with two branches:
- \( 3 \tan\left[ \frac{\pi}{4}(x+2) \right] \) for \( x \leq -1 \)
- \( \sqrt{x^2 + 8} \) for \( x > -1 \)
Each section of a piecewise function must be evaluated separately, and limits need to be considered from both the left and right, as demonstrated in finding \( \lim_{x \rightarrow -1} g(x) \).
Recognizing the structure and evaluating each section individually while considering the appropriate interval helps in fully understanding such functions as exemplified by these kinds of mathematical problems.
Other exercises in this chapter
Problem 54
Evaluate the limits using the limit properties. $$\lim _{x \rightarrow 3} \frac{\frac{1}{x-2}+3 x^{2}}{x^{2}-6+\frac{1}{x-1}}$$
View solution Problem 54
Use a table of values to evaluate the following limits as \(x\) decreases without bound. $$\lim _{x \rightarrow-\infty} \frac{6 x^{2}-x+2}{2 x^{2}+1}$$
View solution Problem 55
Evaluate the limits using the limit properties. $$\lim _{x \rightarrow 3}\left(x^{2}-3\right)^{3}$$
View solution Problem 55
Use a table of values to evaluate the following limits as \(x\) decreases without bound. $$\lim _{x \rightarrow-\infty} \frac{x^{2}+1}{2 x-11}$$
View solution