Problem 10
Question
In Exercises 9-12, find the limit and confirm your answer using the Sandwich Theorem. $$\lim _{x \rightarrow-\infty} \frac{1-\cos x}{x^{2}}$$
Step-by-Step Solution
Verified Answer
The limit of the function \(\frac{1-\cos x}{x^{2}}\) as \(x\) approaches negative infinity is \(0\).
1Step 1: Finding Limit
Firstly, the limit of the given function should be found as \(x\) approaches negative infinity. Due to the complexity of the function, it should be split into two parts to ease calculation:\[\lim_{x \rightarrow-\infty} \frac{1}{x^{2}}- \lim_{x \rightarrow-\infty}\frac{\cos x}{x^{2}}.\]The limit of \(\frac{1}{x^{2}}\) as \(x\) approaches negative infinity is zero. The value of cosine doesn't depend on \(x\) and since \(x\) approaches negative infinity, \(\frac{\cos x}{x^{2}}\) will also tend to zero. This gives limit as \(0-0=0\).
2Step 2: Confirming using Sandwich Theorem
The Sandwich Theorem can now be utilized to verify the obtained answer. It should be noted that \(-1 \leq \cos x \leq 1\). As such, for the negative fraction, the lower limit will be \(\frac{1-1}{x^2}=0\) and the upper limit will be \(\frac{1+1}{x^2}= \frac{2}{x^{2}}\).As \(x\) approaches negative infinity, both the lower and the upper boundaries will tend to zero as well, confirming that the original function must approach zero too, as it is 'sandwiched' between these two values.
Key Concepts
Sandwich TheoremInfinite LimitsCosine FunctionCalculus Problem Solving
Sandwich Theorem
The Sandwich Theorem is a powerful tool in calculus, especially when dealing with limits. It helps find the limit of a function that is "sandwiched" between two other functions. If you can show that two functions both approach the same limit, then any function trapped between them will approach that limit too.
In this exercise, we used the properties of the cosine function, where
In this exercise, we used the properties of the cosine function, where
\(-1 \leq \cos x \leq 1\)to create bounds for the function \(\frac{1-\cos x}{x^2}\). The upper bound was \(\frac{2}{x^2}\) and the lower \(\frac{0}{x^2}\), both tending to zero as \(x\) approaches negative infinity.
- It builds on simple observations about the range of values \(\cos x\) can take.
- Useful in showing limits when direct calculation is complex or impossible.
Infinite Limits
Infinite limits explore what happens to a function as the variable goes towards infinity or negative infinity. In this context, we're interested in how the terms \(\frac{1}{x^2}\) and \(\frac{\cos x}{x^2}\) behave when \(x\) approaches negative infinity.
The first term \(\frac{1}{x^2}\) clearly approaches zero, as any positive number divided by an increasingly large number gets smaller and smaller.
The first term \(\frac{1}{x^2}\) clearly approaches zero, as any positive number divided by an increasingly large number gets smaller and smaller.
- This idea helps simplify many complex expressions.
- Negative infinity just means we're moving further left on the number line—it's not inherently different from positive infinity in these calculations.
Cosine Function
The cosine function, \(\cos x\), is periodic with a range of values between -1 and 1. This property is crucial when applying the Sandwich Theorem.
The periodicity implies that \(\cos x\) doesn't grow or shrink infinitely. Instead, it repeats every cycle.
The periodicity implies that \(\cos x\) doesn't grow or shrink infinitely. Instead, it repeats every cycle.
- This repetition is part of why we can define limits like in the exercise.
- Understanding this aids in visualizing why it doesn't impact the limit as \(x\) approaches infinity.
Calculus Problem Solving
Calculus problem solving often involves breaking down complex problems into simpler pieces. This exercise was initially complicated but splitting it into manageable parts made it easier:
We started by analyzing individual terms. These small steps were key to finding the overall limit.
We started by analyzing individual terms. These small steps were key to finding the overall limit.
- Earlier steps simplify how each part behaves at infinity.
- Combining insights from different limits confirms the solution systematically.
Other exercises in this chapter
Problem 10
In Exercises \(7 - 14 ,\) determine the limit by substitution. Support graphically. $$\lim _ { y \rightarrow 2 } \frac { y ^ { 2 } + 5 y + 6 } { y + 2 }$$
View solution Problem 10
In Exercises \(1-10,\) find the points of continuity and the points of discontinuity of the function. Identify each type of discontinuity. $$y=\ln (x+1)$$
View solution Problem 11
In Exercises \(9-12,\) at the indicated point find (a) the slope of the curve, (b) an equation of the tangent, and (c) an equation of the tangent. (d) Then draw
View solution Problem 11
In Exercises \(7 - 14 ,\) determine the limit by substitution. Support graphically. $$\lim _ { y \rightarrow - 3 } \frac { y ^ { 2 } + 4 y + 3 } { y ^ { 2 } - 3
View solution