Question

graph the function ( y = \tan x ) with the window (-2pi, 2pi \times -10, 10). use the graph to analyze the following limits:
a. ( lim_{x \to \frac{7pi}{2}^+} \tan x )
b. ( lim_{x \to \frac{7pi}{2}^-} \tan x )
c. ( lim_{x \to -\frac{7pi}{2}^+} \tan x )
d. ( lim_{x \to -\frac{7pi}{2}^-} \tan x )
b. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. ( lim_{x \to \frac{7pi}{2}^-} \tan x = ) (simplify your answer)
b. the limit does not exist and is neither ( infty ) nor ( -infty )
c. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. ( lim_{x \to -\frac{7pi}{2}^+} \tan x = ) (simplify your answer)
b. the limit does not exist and is neither ( infty ) nor ( -infty )
d. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

Explanation:

Step1: Recall the properties of the tangent function

The tangent function \( y = \tan x=\frac{\sin x}{\cos x} \) has vertical asymptotes at \( x = \frac{\pi}{2}+n\pi \), where \( n\in\mathbb{Z} \). Let's first find the value of \( \frac{7\pi}{2} \) in terms of the asymptotes. We can write \( \frac{7\pi}{2}=\frac{\pi}{2}+3\pi \), so \( x = \frac{7\pi}{2} \) is a vertical asymptote of \( y = \tan x \).

Step2: Analyze the left - hand limit (\( \lim_{x

ightarrow\frac{7\pi}{2}^{-}}\tan x \))
As \( x \) approaches \( \frac{7\pi}{2} \) from the left (\( x ightarrow\frac{7\pi}{2}^{-} \)), we know that \( \cos x \) approaches \( 0 \) from the positive side (because for \( x \) in the interval \( (\frac{5\pi}{2},\frac{7\pi}{2}) \), \( \cos x>0 \) and as \( x ightarrow\frac{7\pi}{2} \), \( \cos x ightarrow0 \)) and \( \sin x=- 1 \) (since \( \sin(\frac{7\pi}{2})=-1 \) and \( \sin x \) is continuous in the neighborhood of \( \frac{7\pi}{2} \) from the left). So \( \tan x=\frac{\sin x}{\cos x} \), as \( x ightarrow\frac{7\pi}{2}^{-} \), \( \sin x ightarrow - 1 \) and \( \cos x ightarrow0^{+} \), then \( \frac{\sin x}{\cos x} ightarrow-\infty \) (because a negative number divided by a very small positive number goes to negative infinity). Wait, no, let's correct this. Wait, \( \frac{7\pi}{2}=3\pi+\frac{\pi}{2} \). Let's consider the period of \( \tan x \) is \( \pi \). So \( \tan(x)=\tan(x - 3\pi) \). When \( x ightarrow\frac{7\pi}{2}^{-} \), \( x - 3\pi ightarrow\frac{7\pi}{2}-3\pi=\frac{\pi}{2}^{-} \). And we know that \( \lim_{t ightarrow\frac{\pi}{2}^{-}}\tan t=\infty \), so \( \lim_{x ightarrow\frac{7\pi}{2}^{-}}\tan x=\lim_{t ightarrow\frac{\pi}{2}^{-}}\tan t=\infty \) (because \( \tan(x) \) has period \( \pi \), so \( \tan(x)=\tan(x - n\pi) \) for any integer \( n \)).

Step3: Analyze the right - hand limit (\( \lim_{x

ightarrow\frac{7\pi}{2}^{+}}\tan x \))
As \( x \) approaches \( \frac{7\pi}{2} \) from the right (\( x ightarrow\frac{7\pi}{2}^{+} \)), \( x-3\pi ightarrow\frac{\pi}{2}^{+} \). And \( \lim_{t ightarrow\frac{\pi}{2}^{+}}\tan t = -\infty \), so \( \lim_{x ightarrow\frac{7\pi}{2}^{+}}\tan x=\lim_{t ightarrow\frac{\pi}{2}^{+}}\tan t=-\infty \)

Step4: Analyze the two - sided limit (\( \lim_{x

ightarrow\frac{7\pi}{2}}\tan x \))
For the two - sided limit \( \lim_{x ightarrow\frac{7\pi}{2}}\tan x \) to exist, the left - hand limit and the right - hand limit must be equal. But we have \( \lim_{x ightarrow\frac{7\pi}{2}^{-}}\tan x=\infty \) and \( \lim_{x ightarrow\frac{7\pi}{2}^{+}}\tan x = -\infty \), so the two - sided limit does not exist (it is neither \( \infty \) nor \( -\infty \) in the sense of a two - sided limit, and the left and right limits are not equal).

Part b: \( \lim_{x

ightarrow\frac{7\pi}{2}^{-}}\tan x \)
As we analyzed in Step 2, using the periodicity of \( \tan x \) (period \( \pi \)), let \( t=x - 3\pi \). When \( x ightarrow\frac{7\pi}{2}^{-} \), \( t ightarrow\frac{\pi}{2}^{-} \). And \( \lim_{t ightarrow\frac{\pi}{2}^{-}}\tan t=\infty \), so \( \lim_{x ightarrow\frac{7\pi}{2}^{-}}\tan x=\infty \)

Part c: \( \lim_{x

ightarrow\frac{7\pi}{2}^{+}}\tan x \)
As we analyzed in Step 3, let \( t = x-3\pi \). When \( x ightarrow\frac{7\pi}{2}^{+} \), \( t ightarrow\frac{\pi}{2}^{+} \). And \( \lim_{t ightarrow\frac{\pi}{2}^{+}}\tan t=-\infty \), so \( \lim_{x ightarrow\frac{7\pi}{2}^{+}}\tan x = -\infty \)

Part d: \( \lim_{x

ightarrow\frac{7\pi}{2}}\tan x \)
Since \( \lim_{x ightarrow\frac{7\pi}{2}^{-}}\tan x=\infty \) and \( \lim_{x ightarrow\frac{7\pi}{2}^{+}}\tan x=-\infty \), the two - sided limit \( \lim_{x ightarrow\frac{7\pi}{2}}\tan x \) does not exist and is neither \( \infty \) nor \( -\infty \) (in the two - sided sense, and the left and right limits are not equal).

Answers:

Part b:

\( \lim_{x ightarrow\frac{7\pi}{2}^{-}}\tan x=\infty \)

Part c:

\( \lim_{x ightarrow\frac{7\pi}{2}^{+}}\tan x = -\infty \)

Part d:

The limit does not exist and is neither \( \infty \) nor \( -\infty \) (corresponding to option B)

Answer:

s:

Part b:

\( \lim_{x
ightarrow\frac{7\pi}{2}^{-}}\tan x=\infty \)

Part c:

\( \lim_{x
ightarrow\frac{7\pi}{2}^{+}}\tan x = -\infty \)

Part d:

The limit does not exist and is neither \( \infty \) nor \( -\infty \) (corresponding to option B)