QUESTION IMAGE
Question
for the function g whose graph is given, state the value of each quantity, if it exists. (if an answer does not exist, enter dne.)
(a) \\( \lim_{t \to 0^-} g(t) \\)
(b) \\( \lim_{t \to 0^+} g(t) \\)
(c) \\( \lim_{t \to 0} g(t) \\)
(d) \\( \lim_{t \to 2^-} g(t) \\)
(e) \\( \lim_{t \to 2^+} g(t) \\)
(f) \\( \lim_{t \to 2} g(t) \\)
(g) \\( g(2) \\)
(h) \\( \lim_{t \to 4} g(t) \\)
To solve these limit and function value problems, we analyze the graph of \( g(t) \) by examining the left - hand and right - hand behavior as \( t \) approaches a given value, and the actual function value at a point.
Part (a): \( \lim_{t
ightarrow0^{-}}g(t) \)
- Step 1: Understand the left - hand limit
The left - hand limit as \( t
ightarrow0^{-} \) means we look at the values of \( g(t) \) as \( t \) approaches \( 0 \) from the left (values of \( t \) less than \( 0 \)). From the graph, as \( t \) approaches \( 0 \) from the left, the \( y \) - values of the function approach \( - 1 \) (we assume the point on the left - hand curve near \( t = 0 \) has a \( y \) - value of \( - 1 \) based on the grid and the shape of the graph). So, \( \lim_{t
ightarrow0^{-}}g(t)=- 1 \)
Part (b): \( \lim_{t
ightarrow0^{+}}g(t) \)
- Step 1: Understand the right - hand limit
The right - hand limit as \( t
ightarrow0^{+} \) means we look at the values of \( g(t) \) as \( t \) approaches \( 0 \) from the right (values of \( t \) greater than \( 0 \)). From the graph, as \( t \) approaches \( 0 \) from the right, the \( y \) - values of the function approach \( - 1 \) (the right - hand curve near \( t = 0 \) also has a \( y \) - value approaching \( - 1 \)). So, \( \lim_{t
ightarrow0^{+}}g(t)=- 1 \)
Part (c): \( \lim_{t
ightarrow0}g(t) \)
- Step 1: Recall the condition for the existence of a two - sided limit
A two - sided limit \( \lim_{t
ightarrow a}g(t) \) exists if and only if \( \lim_{t
ightarrow a^{-}}g(t)=\lim_{t
ightarrow a^{+}}g(t) \). We found that \( \lim_{t
ightarrow0^{-}}g(t)=-1 \) and \( \lim_{t
ightarrow0^{+}}g(t)=-1 \). Since the left - hand limit and the right - hand limit are equal, \( \lim_{t
ightarrow0}g(t)=-1 \)
Part (d): \( \lim_{t
ightarrow2^{-}}g(t) \)
- **Step 1: Analyze the left - hand limit as \( t
ightarrow2^{-} \)**
The left - hand limit as \( t
ightarrow2^{-} \) means we look at the values of \( g(t) \) as \( t \) approaches \( 2 \) from the left (values of \( t \) less than \( 2 \)). From the graph, as \( t \) approaches \( 2 \) from the left, the \( y \) - values of the function approach \( 2 \) (the left - hand curve near \( t = 2 \) has a \( y \) - value approaching \( 2 \)). So, \( \lim_{t
ightarrow2^{-}}g(t)=2 \)
Part (e): \( \lim_{t
ightarrow2^{+}}g(t) \)
- **Step 1: Analyze the right - hand limit as \( t
ightarrow2^{+} \)**
The right - hand limit as \( t
ightarrow2^{+} \) means we look at the values of \( g(t) \) as \( t \) approaches \( 2 \) from the right (values of \( t \) greater than \( 2 \)). From the graph, as \( t \) approaches \( 2 \) from the right, the \( y \) - values of the function approach \( 1 \) (the right - hand curve near \( t = 2 \) has a \( y \) - value approaching \( 1 \)). So, \( \lim_{t
ightarrow2^{+}}g(t)=1 \)
Part (f): \( \lim_{t
ightarrow2}g(t) \)
- Step 1: Recall the condition for the existence of a two - sided limit
For the two - sided limit \( \lim_{t
ightarrow2}g(t) \) to exist, \( \lim_{t
ightarrow2^{-}}g(t) \) must equal \( \lim_{t
ightarrow2^{+}}g(t) \). We found that \( \lim_{t
ightarrow2^{-}}g(t) = 2 \) and \( \lim_{t
ightarrow2^{+}}g(t)=1 \). Since \( 2
eq1 \), the two - sided limit \( \lim_{t
ightarrow2}g(t) \) does not exist. So, \( \lim_{t
ightarrow2}g(t)=\text{DNE} \)
Part (g): \( g(2) \)
- Step 1: Find the function value at \( t = 2 \)
To find \( g(2) \), we look at the actual point on the graph of \( g(t) \) when \( t = 2 \). From the graph, the filled - in dot (the actual function value) at \( t = 2 \) has a \( y \) -…
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To solve these limit and function value problems, we analyze the graph of \( g(t) \) by examining the left - hand and right - hand behavior as \( t \) approaches a given value, and the actual function value at a point.
Part (a): \( \lim_{t
ightarrow0^{-}}g(t) \)
- Step 1: Understand the left - hand limit
The left - hand limit as \( t
ightarrow0^{-} \) means we look at the values of \( g(t) \) as \( t \) approaches \( 0 \) from the left (values of \( t \) less than \( 0 \)). From the graph, as \( t \) approaches \( 0 \) from the left, the \( y \) - values of the function approach \( - 1 \) (we assume the point on the left - hand curve near \( t = 0 \) has a \( y \) - value of \( - 1 \) based on the grid and the shape of the graph). So, \( \lim_{t
ightarrow0^{-}}g(t)=- 1 \)
Part (b): \( \lim_{t
ightarrow0^{+}}g(t) \)
- Step 1: Understand the right - hand limit
The right - hand limit as \( t
ightarrow0^{+} \) means we look at the values of \( g(t) \) as \( t \) approaches \( 0 \) from the right (values of \( t \) greater than \( 0 \)). From the graph, as \( t \) approaches \( 0 \) from the right, the \( y \) - values of the function approach \( - 1 \) (the right - hand curve near \( t = 0 \) also has a \( y \) - value approaching \( - 1 \)). So, \( \lim_{t
ightarrow0^{+}}g(t)=- 1 \)
Part (c): \( \lim_{t
ightarrow0}g(t) \)
- Step 1: Recall the condition for the existence of a two - sided limit
A two - sided limit \( \lim_{t
ightarrow a}g(t) \) exists if and only if \( \lim_{t
ightarrow a^{-}}g(t)=\lim_{t
ightarrow a^{+}}g(t) \). We found that \( \lim_{t
ightarrow0^{-}}g(t)=-1 \) and \( \lim_{t
ightarrow0^{+}}g(t)=-1 \). Since the left - hand limit and the right - hand limit are equal, \( \lim_{t
ightarrow0}g(t)=-1 \)
Part (d): \( \lim_{t
ightarrow2^{-}}g(t) \)
- **Step 1: Analyze the left - hand limit as \( t
ightarrow2^{-} \)**
The left - hand limit as \( t
ightarrow2^{-} \) means we look at the values of \( g(t) \) as \( t \) approaches \( 2 \) from the left (values of \( t \) less than \( 2 \)). From the graph, as \( t \) approaches \( 2 \) from the left, the \( y \) - values of the function approach \( 2 \) (the left - hand curve near \( t = 2 \) has a \( y \) - value approaching \( 2 \)). So, \( \lim_{t
ightarrow2^{-}}g(t)=2 \)
Part (e): \( \lim_{t
ightarrow2^{+}}g(t) \)
- **Step 1: Analyze the right - hand limit as \( t
ightarrow2^{+} \)**
The right - hand limit as \( t
ightarrow2^{+} \) means we look at the values of \( g(t) \) as \( t \) approaches \( 2 \) from the right (values of \( t \) greater than \( 2 \)). From the graph, as \( t \) approaches \( 2 \) from the right, the \( y \) - values of the function approach \( 1 \) (the right - hand curve near \( t = 2 \) has a \( y \) - value approaching \( 1 \)). So, \( \lim_{t
ightarrow2^{+}}g(t)=1 \)
Part (f): \( \lim_{t
ightarrow2}g(t) \)
- Step 1: Recall the condition for the existence of a two - sided limit
For the two - sided limit \( \lim_{t
ightarrow2}g(t) \) to exist, \( \lim_{t
ightarrow2^{-}}g(t) \) must equal \( \lim_{t
ightarrow2^{+}}g(t) \). We found that \( \lim_{t
ightarrow2^{-}}g(t) = 2 \) and \( \lim_{t
ightarrow2^{+}}g(t)=1 \). Since \( 2
eq1 \), the two - sided limit \( \lim_{t
ightarrow2}g(t) \) does not exist. So, \( \lim_{t
ightarrow2}g(t)=\text{DNE} \)
Part (g): \( g(2) \)
- Step 1: Find the function value at \( t = 2 \)
To find \( g(2) \), we look at the actual point on the graph of \( g(t) \) when \( t = 2 \). From the graph, the filled - in dot (the actual function value) at \( t = 2 \) has a \( y \) - value of \( 1 \). So, \( g(2)=1 \)
Final Answers:
(a) \( \boldsymbol{-1} \)
(b) \( \boldsymbol{-1} \)
(c) \( \boldsymbol{-1} \)
(d) \( \boldsymbol{2} \)
(e) \( \boldsymbol{1} \)
(f) \( \boldsymbol{\text{DNE}} \)
(g) \( \boldsymbol{1} \)