Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 9E from Chapter 2.2 from Stewart's Calculus, 8th Edition.
Problem 9E
Chapter:
Problem:
For the function f whose graph is shown, state the following.
Step-by-Step Solution
Step 1
We are given with a graph of a function. We have to find certain limits.
Step 2: (a)
Limit at x=-7: we can see that at x=-7, the function approaches -$\infty $ from both left and right hand side, :
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = - 7} \,f\left( x \right) = - \infty \]
Step 3: (b)
Limit at x=-3: we can see that at x=-3, the function approaches $\infty $ from both left and right hand side,
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = - 3} \,f\left( x \right) = \infty \]
Step 4: (c)
Limit at x=0: we can see that at x=0, the function approaches $\infty $ from both left and right hand side,
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = 0} \,f\left( x \right) = \infty \]
Step 5: (d)
Limit at $x=6^-$: we can see that at x=6, the function approaches -$\infty $ from left hand side,:
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = 6^-} \,f\left( x \right) = -\infty \]
Step 6: (e)
Limit at $x=6^+$: we can see that at x=6, the function approaches $\infty $ from right hand side,:
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = 6^+} \,f\left( x \right) = \infty \]
Step 7: (f)
At any point, $x=a$ is called an asymptote of the function, if at that point, function approaches either infinity or negative infinity
As we saw from above answers that at x=-7, -3, 0 an 6, the function approaches either negative infinity or infinity:Therefore, the asymptote are \[x = - 7,\,\,x = - 3,\,x = 0,\,\,\,x = 6\]
We are given with a graph of a function. We have to find certain limits.
Step 2: (a)
Limit at x=-7: we can see that at x=-7, the function approaches -$\infty $ from both left and right hand side, :
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = - 7} \,f\left( x \right) = - \infty \]
Step 3: (b)
Limit at x=-3: we can see that at x=-3, the function approaches $\infty $ from both left and right hand side,
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = - 3} \,f\left( x \right) = \infty \]
Step 4: (c)
Limit at x=0: we can see that at x=0, the function approaches $\infty $ from both left and right hand side,
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = 0} \,f\left( x \right) = \infty \]
Step 5: (d)
Limit at $x=6^-$: we can see that at x=6, the function approaches -$\infty $ from left hand side,:
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = 6^-} \,f\left( x \right) = -\infty \]
Step 6: (e)
Limit at $x=6^+$: we can see that at x=6, the function approaches $\infty $ from right hand side,:
Therefore, \[\mathop {{\rm{lim}}\,}\limits_{x = 6^+} \,f\left( x \right) = \infty \]
Step 7: (f)
At any point, $x=a$ is called an asymptote of the function, if at that point, function approaches either infinity or negative infinity
As we saw from above answers that at x=-7, -3, 0 an 6, the function approaches either negative infinity or infinity:Therefore, the asymptote are \[x = - 7,\,\,x = - 3,\,x = 0,\,\,\,x = 6\]