Question

Find (a) \(\lim _{x \rightarrow 0+} \frac{1}{x^{3}}\) (b) \(\lim _{x \rightarrow 0-} \frac{1}{x^{3}}\) (c) \(\lim _{x \rightarrow 0+} \frac{1}{x^{6}}\) (d) \(\lim _{x \rightarrow 0-} \frac{1}{x^{6}}\) (e) \(\lim _{x \rightarrow x_{0}+} \frac{1}{\left(x-x_{0}\right)^{2 k}}\) (f) \(\lim _{x \rightarrow x_{0}-} \frac{1}{\left(x-x_{0}\right)^{2 k+1}}\) \((k=\) positive integer)

Short answer

Question: Find the limits of the following functions: a) \(\lim _{x \rightarrow 0+} \frac{1}{x^{3}}\) b) \(\lim _{x \rightarrow 0-} \frac{1}{x^{3}}\) c) \(\lim _{x \rightarrow 0+} \frac{1}{x^{6}}\) d) \(\lim _{x \rightarrow 0-} \frac{1}{x^{6}}\) e) \(\lim _{x \rightarrow x_{0}+} \frac{1}{(x-x_{0})^{2k}}\) (k is a positive integer) f) \(\lim _{x \rightarrow x_{0}-} \frac{1}{(x-x_{0})^{2k+1}}\) (k is a positive integer) Answer: a) \(\lim _{x \rightarrow 0+} \frac{1}{x^{3}} = \infty\) b) \(\lim _{x \rightarrow 0-} \frac{1}{x^{3}} = -\infty\) c) \(\lim _{x \rightarrow 0+} \frac{1}{x^{6}} = \infty\) d) \(\lim _{x \rightarrow 0-} \frac{1}{x^{6}} = \infty\) e) \(\lim _{x \rightarrow x_{0}+} \frac{1}{(x-x_{0})^{2k}} = \infty\) f) \(\lim _{x \rightarrow x_{0}-} \frac{1}{(x-x_{0})^{2k+1}} = -\infty\)

Step-by-step solution

Problem (a)

We are asked to find \(\lim _{x \rightarrow 0+} \frac{1}{x^{3}}\). As x approaches 0 from the positive side, we can see that the denominator gets smaller and smaller. This means the overall fraction will get larger and larger, approaching positive infinity. Therefore, the limit is: $$\lim _{x \rightarrow 0+} \frac{1}{x^{3}} = \infty$$

Problem (b)

We are asked to find \(\lim _{x \rightarrow 0-} \frac{1}{x^{3}}\). As x approaches 0 from the negative side, the denominator is still getting smaller, but now it is a negative number raised to the power of 3, which stays negative. Therefore, the limit is: $$\lim _{x \rightarrow 0-} \frac{1}{x^{3}} = -\infty$$

Problem (c)

We are asked to find \(\lim _{x \rightarrow 0+} \frac{1}{x^{6}}\). As x approaches 0 from the positive side, we can see that the denominator gets smaller and smaller. However, since the exponent is even, the denominator is always positive, making the overall fraction positive. Therefore, the limit is: $$\lim _{x \rightarrow 0+} \frac{1}{x^{6}} = \infty$$

Problem (d)

We are asked to find \(\lim _{x \rightarrow 0-} \frac{1}{x^{6}}\). As x approaches 0 from the negative side, the denominator still gets smaller but, since the exponent is even, the denominator is positive. Therefore, the limit is: $$\lim _{x \rightarrow 0-} \frac{1}{x^{6}} = \infty$$

Problem (e)

We are asked to find \(\lim _{x \rightarrow x_{0}+} \frac{1}{\left(x-x_{0}\right)^{2 k}}\), where k is a positive integer. Since the exponent is even, as x approaches \(x_0\) from the positive side, the denominator gets smaller and smaller, resulting in a larger fraction. Therefore, the limit is: $$\lim _{x \rightarrow x_{0}+} \frac{1}{\left(x-x_{0}\right)^{2 k}} = \infty$$

Problem (f)

We are asked to find \(\lim _{x \rightarrow x_{0}-} \frac{1}{\left(x-x_{0}\right)^{2 k+1}}\), where k is a positive integer. Since the exponent is odd, as x approaches \(x_0\) from the negative side, the denominator gets smaller and becomes negative. Therefore, the limit is: $$\lim _{x \rightarrow x_{0}-} \frac{1}{\left(x-x_{0}\right)^{2 k+1}} = -\infty$$

Key concepts

Understanding One-Sided Limits

Grasping the concept of one-sided limits is fundamental in the study of calculus. Imagine approaching a point on a graph from one specific direction—either from the left or the right. When we do this, we're looking at a one-sided limit. Think of it as sneaking up on a point along the x-axis from one side without crossing over it.

For instance, when we consider the limit as 'x' approaches 0 from the positive side, denoted as \(x \rightarrow 0+\), we are only concerned with values of 'x' that are greater than 0, no matter how small. Similarly, when we approach 0 from the left, denoted as \(x \rightarrow 0-\), we consider values just less than 0. The behavior can be quite different when we look at the limits from these individual perspectives, as certain functions may behave erratically at a point but can be well-behaved when approached from one side only.

Limit Approaching Infinity

The limit approaching infinity deals with understanding what happens to a function's value as the input grows without bound or, conversely, as it decreases without bound toward a negative infinity. Take for example the expression \(1/x^3\). As 'x' gets closer to zero from the positive side, the expression grows larger because the value of the denominator decreases significantly, while the numerator remains constant. So we say that the limit of \(1/x^3\) as 'x' approaches 0 from the right is infinity, symbolically: \( \lim _{x \rightarrow 0+} 1/x^3 = \infty \). This concept of infinity in limits helps us understand the behavior of functions that don't settle on a particular value but instead grow without bound.

Even and Odd Exponents in Limits

When dealing with even and odd exponents in limits, we need to consider the parity of the exponent—that is, whether it is even or odd. The general rule is that if we have a function with a variable raised to an even power, like \(x^6\), it will always yield a positive result irrespective of whether 'x' is positive or negative. That's why the limits as 'x' approaches 0 from either direction for \(1/x^6\) will be infinity.

On the other hand, variables raised to odd powers, such as \(x^3\), retain the sign of the variable. This means that \(1/x^3\) approaches positive infinity as 'x' approaches zero from the right but approaches negative infinity as 'x' goes to zero from the left side. It's this change in sign that can dramatically affect the behavior of a function near points where the denominator goes to zero, and it’s vital to distinguish this when solving limit problems.