Key Types of Questions: Limits and Derivatives - Common Calculus Exam Problem Analysis

The concept of limits is the foundation of calculus and serves as the crucial bridge connecting differentiation and integration. This unit will focus on the computation of limits and how to use limits to find derivatives, differential functions, and asymptotes.


Precise Definition of Limits

Do not underestimate this concept. In high school, we learn an intuitive definition of limits, but the college definition is different and often a focal point for professors. Here’s a review of the precise definition of limits:

We write\[\lim\limits_{x\to a}f\left(x\right)=L\]

if for every number \(\delta \gt 0\) there is a number \(\varepsilon \gt 0\) such that

\(\bbox[#e0ebeb,5px,border:3px solid #b3f0ff]{\text{if}~0<\left|x-a\right|<\delta~~~ \text{then}~~  \left|f(x)-L\right|<\varepsilon}\)


Let's dive into an example:

Let \(A=\{0.6,0.7,0.8,0.9\},\) find the largest number, \(\delta\), in \(A\) such that \(\left|\sqrt{4x+5}-3\right|\lt0.6\), whenever \(\left|x-1\right|\lt\delta\).

(A) 0.6 ,      (B) 0.7 ,      (C) 0.8 ,      (D) 0.9 .

From the left-hand side, we get:

\begin{aligned} &−0.6\lt\sqrt{4x+5}−3\lt0.6\\ &\Rightarrow2.4\lt\sqrt{4x+5}\lt3.6\\ &\Rightarrow5.76\lt\sqrt{4x+5}\lt12.96\\ &\Rightarrow0.76\lt4x\lt7.96\\ &\Rightarrow0.19\lt x\lt1.99\\ &\Rightarrow−0.81\lt x−1\lt0.99 \end{aligned}

This part should be straightforward, but the next step is critical: should we choose 0.81 or 0.99? Looking back at the question, we need the largest number, which is 0.99. However, we should choose the largest number that doesn't exceed 0.99, so (D) 0.9.

If you chose 0.99, you would be incorrect. Here's why:

\begin{array}{rrr} -0.81\overset{0.81}{\longleftrightarrow} x\overset{0.99}{\longleftrightarrow}0.99\\ \hline \end{array}

If \(\delta=0.99\), \(x\) can go as low as -0.9, exceeding the -0.81 limit. Therefore, although the problem asks for the largest number, the correct answer is the minimum of 0.81 and 0.99, which is 0.81. Thus, choose (C) 0.8.

Limits of Composite Functions

Besides arithmetic operations, limits of composite functions are significant. Here's the crucial formula:

If \(f\) is continuous at \(b\) and \(\lim\limits_{x\to a}g(x) = b\), then

\(\bbox[#e0ebeb,5px,border:3px solid #b3f0ff]{\lim\limits_{x\to a}f(g(x)) = f\left(\lim\limits_{x\to a}g(x)\right)}\)

Don't believe it? Let's see an example:

Evaluate the limit of \(\lim\limits_{{x \to 0}} \left( \dfrac{1}{\ln(x+1)} - \dfrac{x+1}{x} \right)^2.\)

(A) 0; (B) \(\dfrac{1}{4}\); (C) 1; (D) nonexistent.

Using the above formula, we only need to calculate

\[\lim\limits_{{x \to 0}} \left( \dfrac{1}{\ln(x+1)} - \dfrac{x+1}{x} \right) = \lim\limits_{{x \to 0}} \dfrac{x - (x+1) \ln(x+1)}{x \ln(x+1)}\]

Then square the result; no need to square first and then find the limit, which would waste time and effort.

This formula can also be used in some proofs, particularly those involving limits of indeterminate forms.

Squeeze Theorem

The Squeeze Theorem is both loved and hated in high school and college. It's loved because if you can identify the upper and lower bound functions, you can solve problems quickly. It's hated because sometimes you need a "sixth sense" to see them. Generally, the Squeeze Theorem is only used as a last resort. However, there are some tips that make it easier to spot the bounds. Let's look at an example:

\(\lim\limits_{x\to\infty}\left(\dfrac{\ln{x}}{x^e}\dfrac{x^\pi}{e^x}\right) =\)

(A) \(\infty\) ,    (B) \(\pi\) ,   (C) \(\dfrac{e}{\pi}\) ,   (D) \(0\) .

One key to solving problems using the Squeeze Theorem is reverse thinking, which means working backwards from the result. More precisely, it involves guessing the possible upper and lower bound functions based on the four given options.

As a kid, we were taught to guess 0 when we didn't know the answer in math. If the answer is 0, the denominator should approach infinity. Both \(x^e\) and \(e^x\) fit this condition, but choosing \(e^x\) is usually more convenient (since it remains the same after differentiation). So we get:

\[\dfrac{1}{e^x}\lt \dfrac{\ln{x}}{x^e}\dfrac{x^\pi}{e^x}\lt \dfrac{x^\pi}{e^x}\]

Here, we don't need to use L'Hôpital's rule; it's intuitive to get:

\[\lim\limits_{x\to\infty}\dfrac{1}{e^x} = \lim\limits_{x\to\infty}\dfrac{x^\pi}{e^x} = 0\]

So, by the Squeeze Theorem, we know:

\[\lim\limits_{x\to\infty}\left(\dfrac{\ln{x}}{x^e}\dfrac{x^\pi}{e^x}\right) = 0\]

Definition of Derivatives

Derivatives and differential functions are defined by limits, but after learning the differentiation formulas, many people forget their original definitions. Sometimes, returning to the definition makes solving certain problems easier:

\(\text{Let } f(x) = \dfrac{\tan 2x \cdot \cos^{-1} x + \ln(1+x)}{3 \sec^3 x + x^3 \sin^{-1} x}. \text{ Find } f'(0).\)

Using differentiation formulas here is a trap, as the whole expression would become extremely long. So, we need to use the definition of the derivative:

\[\lim_{{x \to 0}} \frac{f(x) - f(0)}{x - 0}  = \lim_{{x \to 0}} \frac{1}{3 \sec^2 x + x^3 \sin^{-1} x} \cdot \left( \frac{2 \tan 2x}{2x} \cos^{-1} x + \frac{\ln(1+x)}{x} \right)\]

\[= \frac{1}{3} \left( 1 \cdot 2 \cdot \frac{\pi}{2} + 1 \right) = \frac{\pi + 1}{3}\]


Finding Asymptotes

Limits can be used to find derivatives and differential functions, but with differentiation formulas available, they are not often used for this purpose. However, limits are essential for finding asymptotes. Although we haven't discussed this in detail yet, I want to first prove a useful formula for finding asymptotes using the properties of limits (only proving for positive infinity, the same applies for negative infinity):

A function \(f(x)\) has a slant asymptote \(y=ax+b\) if \[\lim\limits_{x\to\pm\infty}\left[f(x)-(ax+b)\right] = 0\]

then \[\bbox[#e0ebeb,5px,border:3px solid #b3f0ff]{\left\{\begin{array}{l}a = \lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x} \\b = \lim\limits_{x\to\pm\infty}\left(f(x)-ax\right) \end{array}\right. }\]


To find \( a \), consider the expression for \( \dfrac{f(x)}{x} \):

\[\dfrac{f(x)}{x} = \dfrac{ax + b + (f(x) - (ax + b))}{x} = a + \dfrac{b}{x} + \dfrac{f(x) - (ax + b)}{x}\]

Taking the limit as \( x \to \infty \):

\[\lim_{x \to \infty} \dfrac{f(x)}{x} = \lim_{x \to \infty} \left( a + \dfrac{b}{x} + \dfrac{f(x) - (ax + b)}{x} \right)\]

Since \( \lim\limits_{x \to \infty} \dfrac{b}{x} = 0 \) and \( \lim\limits_{x \to \infty} \dfrac{f(x) - (ax + b)}{x} = 0 \) (because \( f(x) - (ax + b) \to 0 \) by definition of the slant asymptote), we get:

\[\lim\limits_{x \to \infty} \dfrac{f(x)}{x} = a ~~.\]


To find \( b \), consider the expression for \( f(x) - ax \):

\[f(x) - ax = (ax + b + (f(x) - (ax + b))) - ax = b + (f(x) - (ax + b))\]

Taking the limit as \( x \to \infty \):

\[\lim_{x \to \infty} (f(x) - ax) = \lim\limits_{x \to \infty} \left( b + (f(x) - (ax + b)) \right)\]

Since \( \lim\limits_{x \to \infty} (f(x) - (ax + b)) = 0 \), we get:

\[\lim\limits_{x \to \infty} (f(x) - ax) = b ~~.\]


Therefore, if \( f(x) \) has a slant asymptote \( y = ax + b \) when \( x \to \infty \), then \( a = \lim\limits_{x \to \infty} \dfrac{f(x)}{x} \) and \( b = \lim\limits_{x \to \infty} (f(x) - ax) \).

Understanding this formula, it becomes much easier to find asymptotes using methods like L'Hôpital's rule.

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