Limit Using Expansion: Formula, Series

Important Series Expansions in Limits

When solving limits that lead to indeterminate forms, having a strong command over the important expansions in limits is essential. These expansions are derived from the Maclaurin or Taylor series and represent functions as infinite polynomials. Using these expansions in limits, we can simplify complex expressions, especially as $x \to 0$.

Below is a list of the most important series expansions in limits, which are frequently used in calculus and competitive exams:

Standard Series Expansions:

• Exponential Function
  $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$

• Logarithmic Function
  $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \quad \text{for } |x| < 1$

• Sine Function
  $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$

• Cosine Function
  $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

• Tangent Function
  $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \cdots \quad \text{for small } x$

• Binomial Expansion (for any real number $n$)
  $(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \cdots$

These expansions of all functions are valid in the neighbourhood of $x = 0$ and are widely applied in solving limits, especially when algebraic simplification is not possible. They also provide accurate approximations when $x$ is very small.

How These Expansions Help in Limits:

• Turn indeterminate forms into algebraic expressions that are easy to evaluate.

• Allow cancellation of terms in the numerator and denominator.

• Avoid the need for derivatives or L’Hôpital’s Rule.

• Clarify the behaviour of functions near $x = 0$.

How to Apply Series Expansion in Limits

Applying the limit using the series expansion technique involves a structured process that transforms complex or indeterminate limit expressions into solvable algebraic forms. Instead of relying on algebraic manipulation or L’Hôpital’s Rule, this method uses limits expansion formulas to replace functions with their equivalent series.

Here is a step-by-step guide to apply the series expansion in limits:

Step 1: Identify the Indeterminate Form

Check if the given limit leads to a form like $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, or $\infty - \infty$. These forms signal that the expression can be simplified using series.

Step 2: Choose the Correct Series Expansion

Select the appropriate limits expansion formula based on the functions involved. Common choices include expansions for $\sin x$, $\cos x$, $\ln(1 + x)$, $e^x$, $\tan x$, and $(1 + x)^n$.

Step 3: Substitute and Simplify

Replace the original function(s) with their series expansion (typically up to the lowest necessary degree), and simplify the expression by cancelling or combining terms. Then compute the resulting limit.

Example 1: Evaluate $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Step-by-step using series expansion:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

So, $e^x - 1 - x = \left(1 + x + \frac{x^2}{2} + \cdots \right) - 1 - x = \frac{x^2}{2} + \cdots$

Now, $\frac{e^x - 1 - x}{x^2} = \frac{\frac{x^2}{2} + \cdots}{x^2} = \frac{1}{2} + \cdots$

Therefore, $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \frac{1}{2}$

Example 2: Evaluate $\lim_{x \to 0} \frac{\ln(1 + x) - x + \frac{x^2}{2}}{x^3}$

Use the expansion:
$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$

So, $\ln(1 + x) - x + \frac{x^2}{2} = \frac{x^3}{3} - \cdots$

Then, $\frac{\ln(1 + x) - x + \frac{x^2}{2}}{x^3} = \frac{\frac{x^3}{3} - \cdots}{x^3} = \frac{1}{3} + \cdots$

Therefore, $\lim_{x \to 0} \frac{\ln(1 + x) - x + \frac{x^2}{2}}{x^3} = \frac{1}{3}$

Limits Using Series Expansions

Several important limits in calculus can be directly evaluated using known limit expansion formulas. These limits are foundational and often appear in derivative definitions, continuity proofs, and advanced limit problems. Understanding these helps in mastering the technique of limit using series expansion.

Here are some standard results:

$\text{1. } \lim_{x \to 0} \frac{\sin x}{x} = 1$

$\text{2. } \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

$\text{3. } \lim_{x \to 0} \frac{\tan x}{x} = 1$

$\text{4. } \lim_{x \to 0} \frac{e^x - 1}{x} = 1$

$\text{5. } \lim_{x \to 0} \frac{\log(1 + x)}{x} = 1$

These results are derived from the expansion in limits using the Maclaurin series and are frequently used to simplify expressions involving indeterminate forms.

Why Use Series Expansion in Limits?

• It simplifies indeterminate expressions that are otherwise hard to evaluate.

• It offers an alternative to L’Hôpital’s Rule with deeper insight.

• It provides exact values using limit expansion series without relying on approximation.

• It is especially useful for small-angle approximations and competitive exams.

Solved Examples Based On Limits Using Expansion

Example 1: Find the limit $\lim _{x \rightarrow 0} \frac{2^x-1}{(1+x)^{1 / 2}-1}=$

1) $\log 2$

2) $\log 4$

3) $\log{\sqrt{2}}$

4) None of these

Solution:

Using the expansion of some functions is one of the easier methods to finding the limit of expression. The following expansion formulas which is also known as Taylor series, are very useful in evaluating various limits.

As we know,

$a^x=1+x\left(\log _e a\right)+\frac{x^2}{2!}\left(\log _e a\right)^2+\frac{x^3}{3!}\left(\log _e a\right)^3+\ldots \ldots \ldots$

$(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots \ldots$

$\lim _{x \rightarrow 0} \frac{2^x-1}{(1+x)^{1 / 2}-1}=\frac{x \log _e 2+\frac{x^2}{2!}\left(\log _e 2\right)^2+\frac{x^3}{3!}\left(\log _e 2\right)^3+\ldots \ldots \ldots}{\frac{1}{2} x+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!} x^2+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!} x^3+\ldots \ldots \ldots}$

$=2 \log _e 2$

$=\log _e 4$

Hence, the answer is option 2.

Example 2: Find the limit $\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=$

1) $\frac{m}{n}$

2) $\frac{n}{m}$

3) $\frac{m^2}{n^2}$

4) $\frac{n^2}{m^2}$

Solution:

As we know,

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots \ldots$

$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \frac{\frac{(m x)^2}{2!}-\frac{(m x)^4}{4!} \cdots}{\frac{(n x)^2}{2!}-\frac{(m x)^4}{4!} \cdots}=\frac{m^2}{n^2}$

Hence, the answer is the option 3.

Example 3: Find limit $\lim _{x \rightarrow 0} \frac{\sin m x}{\sin n x}=$

1) 0

2) 1

3) $\frac{m}{n}$

4) $\frac{n}{m}$

Solution:

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots \ldots$

$\lim _{x \rightarrow 0} \frac{\sin m x}{\sin n x}=\lim _{x \rightarrow 0} \frac{m x-\frac{(m x)^3}{3!}+\frac{(m x)^5}{5!}-\ldots \ldots}{n x-\frac{(n x)^3}{3!}+\frac{(n x)^5}{5!}-\ldots \ldots}$

Hence, the answer is the option 3.

Example 4: Without using the method of L' Hospital, find the limit.

$\lim _{x \rightarrow 0} \frac{(\sqrt[4]{a})^x+\log _e(1+x)-\left(e^2\right)^x}{x}$

1) $\frac{\ln a}{4}$

2) $\frac{\ln a}{4}+1$

3) $\frac{\ln a}{4}-2$

4) $\frac{\ln a}{4}-1$

Solution:

Now, evaluate the following limit using the above formula.

$\lim _{x \rightarrow 0} \frac{(\sqrt[4]{a})^x+\log _a(1+x)-\left(e^2\right)^x}{x}$

$=\lim _{x \rightarrow 0} \frac{a^{\frac{x}{4}}-1+\log _e(1+x)-e^{2 x}+1}{x}$

$=\frac{1}{4} \times \lim _{x \rightarrow 0} \frac{a^{\frac{x}{4}}-1}{\frac{x}{4}}+\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{x}-2 \times \lim _{x \rightarrow 0} \frac{e^{2 x}-1}{2 x}$

$=\frac{1}{4} \times \ln a+1-2 \times 1$

$=\frac{\ln a}{4}-1$

Hence, the answer is the option 4.

Example 5: Let the $r$ th term $t_r$ of a series is given by $t_r=\frac{r}{1+r^2+r^4}$. Then $\lim _{n \rightarrow \infty} \sum_{r=1}^n t_r$ is

1) $\frac{1}{4}$

2) 1

3) $\frac{1}{2}$

4) None of these

Solution:

$\mathrm{t}_{\mathrm{r}}=\frac{1}{2} \cdot \frac{2 \mathrm{r}}{\left(\mathrm{r}^2+1\right)^2-\mathrm{r}^2}=\frac{1}{2}\left\{\frac{1}{\mathrm{r}^2-\mathrm{r}+1}-\frac{1}{\mathrm{r}^2+\mathrm{r}+1}\right\}$

$=\frac{1}{2}\left\{\frac{1}{\mathrm{r}(\mathrm{r}-1)+1}-\frac{1}{(\mathrm{r}+1) \mathrm{r}+1}\right\}$

$\therefore \quad \sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{t}_{\mathrm{r}}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{2}\{\mathrm{f}(\mathrm{r})-\mathrm{f}(\mathrm{r}+1)\}, \text { where } \mathrm{f}(\mathrm{r})=\frac{1}{\mathrm{r}(\mathrm{r}-1)+1}$

$=\frac{1}{2}\{\mathrm{f}(1)-\mathrm{f}(\mathrm{n}+1)\}=\frac{1}{2}\left\{1-\frac{1}{(\mathrm{n}+1) \mathrm{n}+1}\right\} \rightarrow \frac{1}{2} \text { as } \mathrm{n} \rightarrow \infty$

Hence, the answer is the option 3.

List of Topics related to Limits using Expansion

Explore the key topics connected to solving limits using series expansions. From foundational limit formulas to trigonometric and algebraic limit rules, these topics offer a structured path to mastering limit problems with expansion techniques. Use the links below to access detailed formulas and examples for each topic.

NCERT Resources

Explore curated NCERT study materials for "Limits and Derivatives", including comprehensive notes, accurate solutions, and exemplar problems with detailed explanations. These resources are essential for mastering the fundamentals and aligning with the CBSE curriculum. Here is a table with direct links to Careers360's NCERT resources for "Limits and Derivatives":

Practice Questions based on Limits using Expansion

Strengthen your understanding of limits through expansion techniques with targeted MCQs. These practice questions are designed to test your application of series expansions, standard limits, and algebraic manipulations in limit problems. Perfect for reinforcing concepts and preparing for competitive exams.

Limit Using Expansion - Practice Question MCQ

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