In exploring the concept of limits within calculus, it's critical to understand that not all functions have limits at every point. This section delves into specific scenarios where limits do not exist: when a function becomes unbounded, oscillates without approaching a specific value, or behaves differently when approached from the left compared to the right. Through mathematical equations and graphical examples, we aim to illustrate these conditions, reinforcing the understanding that limits are an essential yet sometimes elusive aspect of function analysis.

Unbounded Behavior

• Definition: A limit does not exist if the function approaches infinity or negative infinity as $x$ approaches a certain value.
• Example:

$\lim_{x \to 0} \dfrac{1}{x^2}$

• Calculation:

$ \begin{aligned} &\text{As } x \to 0^+, \frac{1}{x^2} \to \infty \\ &\text{As } x \to 0^-, \frac{1}{x^2} \to \infty \end{aligned}$<p></p><ul><li><strong>Conclusion</strong>: Since the function becomes infinitely large as$x$approaches 0, the limit does not exist.</li></ul><h2 id="oscillating-behavior"><strong>Oscillating Behavior</strong></h2><ul><li><strong>Definition</strong>: A limit does not exist if the function oscillates between values as (x) approaches a certain point without settling on a single value.</li><li><strong>Example</strong>: </li></ul>$\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$<p></p><ul><li><strong>Calculation</strong>: </li></ul>$ \begin{aligned} &\text{As } x \to 0, \sin\left(\frac{1}{x}\right) \text{ oscillates between } -1 \text{ and } 1 \\ &\text{No single limit value can be identified.} \end{aligned} $<p></p><ul><li><strong>Conclusion</strong>: The oscillating nature of$\sin\left(\frac{1}{x}\right)$near$x = 0$means the limit does not exist.</li></ul><h2 id="behavior-differing-by-direction"><strong>Behavior Differing by Direction</strong></h2><ul><li><strong>Definition</strong>: A limit does not exist if the left-hand limit and right-hand limit as$x$approaches a certain value are not equal.</li><li><strong>Example</strong>: </li></ul>$\lim_{x \to 0} \dfrac{|x|}{x}$<p></p><ul><li><strong>Calculation</strong>: </li></ul>$ \begin{aligned} &\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1 \\ &\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1 \end{aligned}$<p></p><ul><li><strong>Conclusion</strong>: Since$\lim{x \to 0^+} \neq \lim{x \to 0^-}$, the limit as$x \to 0$does not exist.</li></ul><h2 id="graphical-examples"><strong>Graphical Examples</strong></h2><p><strong>Example 1</strong>: Graph of$y = \dfrac{1}{x^2}$.</p><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/e528fe8a-1dc0-4288-abce-1ff667ae1bf6-file.png" alt="Graph of y = 1/x^2" style="width: 500px; height: 441px" width="500" height="441"><ul><li><strong>Features</strong>: The graph shows that as$x$approaches 0 from both sides,$y$becomes infinitely large, indicating an unbounded behavior.</li></ul><p></p><p><strong>Example 2</strong>: Graph of$y = \sin\left(\dfrac{1}{x}\right)$.</p><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/e6c92e81-ab35-406a-b800-ddd06b8929cc-file.png" alt="Graph of y = sin (1/x)" style="width: 600px; height: 453px" width="600" height="453"><ul><li><strong>Features</strong>: As$x$approaches 0, the graph oscillates infinitely between -1 and 1 without approaching a specific value, illustrating oscillating behavior.</li></ul><p></p><p><strong>Example 3</strong>: Graph of$y = \dfrac{|x|}{x}$.</p><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/0db29cdb-5fd3-4b18-b639-9b41f115edd2-file.png" alt="Graph of y = |x|/x" style="width: 600px; height: 453px" width="600" height="453"><ul><li><strong>Features</strong>: The graph illustrates a jump discontinuity at$x = 0$, where the left-hand limit is -1 and the right-hand limit is 1, demonstrating behavior differing by direction.</li></ul><h2 id="practice-questions">Practice Questions</h2><h3>Question 1</h3><p>Determine if the limit exists for$f(x) = \dfrac{2x}{|x|}$as$x$approaches 0.</p><h3>Question 2</h3><p>Evaluate the limit of$g(x) = x \sin\left(\dfrac{1}{x}\right)$as$x$approaches 0.</p><h3>Question 3</h3><p>Investigate the existence of the limit for$h(x) = \dfrac{1}{x^2 - 1}$as$x$approaches 1.</p><h3>Question 4</h3><p>Determine the limit of$f(x) = \cos\left(\pi x\right)$as$x$approaches 2 from the left and the right.</p><h2 id="solutions-to-practice-questions">Solutions to Practice Questions</h2><h3>Solution to Question 1</h3><ul><li><strong>Left-hand Limit</strong>: </li></ul>$\lim{x \to 0^-} \frac{2x}{|x|} = \lim{x \to 0^-} \frac{2x}{-x} = -2$<ul><li><strong>Right-hand Limit</strong>: </li></ul>$\lim{x \to 0^+} \frac{2x}{|x|} = \lim{x \to 0^+} \frac{2x}{x} = 2$<ul><li><strong>Conclusion</strong>: Since the left-hand limit$-2$is not equal to the right-hand limit$2$, the limit does not exist.</li></ul><p></p><h3>Solution to Question 2</h3><ul><li><strong>Approach</strong>: Utilizing the Squeeze Theorem.</li><li><strong>Calculation</strong>: Since$\sin\left(\frac{1}{x}\right)$oscillates between$-1$and$1$as$x$approaches 0, and$x$multiplies the sine function, we evaluate the behavior around 0:$-x \leq x \sin\left(\frac{1}{x}\right) \leq x$, As$x \to 0$both bounds$-x$and$x$approach 0.</li><li><strong>Conclusion</strong>: By the Squeeze Theorem, since both bounds go to 0, the limit of$g(x)$as$x \to 0$also goes to 0. The limit exists and is 0.</li></ul><p></p><h3>Solution to Question 3</h3><ul><li><strong>Calculation</strong>: </li></ul>$\lim_{x \to 1} \dfrac{1}{x^2 - 1}$<p></p><p> Observing that$x^2 - 1 = (x+1)(x-1)$, as$x$approaches 1, the denominator approaches 0, which suggests the function's value becomes infinitely large or small.</p><ul><li><strong>Conclusion</strong>: The function$h(x)$becomes unbounded near$x = 1$, thus the limit does not exist.</li></ul><p></p><h3>Solution to Question 4</h3><ul><li><strong>Left-hand Limit</strong>: </li></ul>$\lim_{x \to 2^-} \cos\left(\pi x\right) = \cos(2\pi) = 1$<ul><li><strong>Right-hand Limit</strong>: </li></ul>$\lim_{x \to 2^+} \cos\left(\pi x\right) = \cos(2\pi) = 1$<ul><li><strong>Conclusion</strong>: The limit as$x$ approaches 2 exists and is 1, as both the left-hand and right-hand limits are equal, confirming the existence and value of the limit.