examples of limit computations

by Kurious Fox

Here are a few more examples of limit computations involving various techniques:

Example 1: Basic Limit

Find the limit: \lim_{x \to 2} (3x + 5)

Solution:

This is a basic limit where we can directly substitute x = 2:
\lim_{x \to 2} (3x + 5) = 3(2) + 5 = 6 + 5 = 11

Example 2: Limit Involving a Rational Function

Find the limit:
\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Solution:

First, notice that the expression is indeterminate of the form \frac{0}{0} when x = 3. To resolve this, factor the numerator:
x^2 - 9 = (x - 3)(x + 3)
So the limit becomes:
\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}

We can cancel the x - 3 terms:
\lim_{x \to 3} (x + 3)

Now, substitute x = 3:
3 + 3 = 6

Example 3: Limit Involving Infinity

Find the limit:
\lim_{x \to \infty} \frac{5x^2 + 3x}{2x^2 - x}

Solution:

To find the limit as x \to \infty, we divide the numerator and the denominator by the highest power of x in the denominator, which is x^2:
\lim_{x \to \infty} \frac{\frac{5x^2}{x^2} + \frac{3x}{x^2}}{\frac{2x^2}{x^2} - \frac{x}{x^2}} = \lim_{x \to \infty} \frac{5 + \frac{3}{x}}{2 - \frac{1}{x}}

As x \to \infty, \frac{3}{x} \to 0 and \frac{1}{x} \to 0, so the limit simplifies to:
\frac{5 + 0}{2 - 0} = \frac{5}{2}

Example 4: Limit Involving Trigonometric Functions

Find the limit:
\lim_{x \to 0} \frac{\sin(x)}{x}

Solution:

This is a well-known limit and a fundamental result in calculus:
\lim_{x \to 0} \frac{\sin(x)}{x} = 1

Example 5: L’Hôpital’s Rule

Find the limit:
\lim_{x \to 0} \frac{e^x - 1}{x}

Solution:

This limit is indeterminate of the form \frac{0}{0}. We can use L’Hôpital’s Rule, which states that if \lim_{x \to c} \frac{f(x)}{g(x)} is indeterminate, then it equals \lim_{x \to c} \frac{f'(x)}{g'(x)} if the latter limit exists.

Here, f(x) = e^x - 1 and g(x) = x. Taking the derivatives:
f'(x) = e^x \quad \text{and} \quad g'(x) = 1

Applying L’Hôpital’s Rule:
\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1

Example 6: Squeeze Theorem

Find the limit:
\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)

Solution:

The function \sin\left(\frac{1}{x}\right) oscillates between -1 and 1 for all x, so we have:
-1 \leq \sin\left(\frac{1}{x}\right) \leq 1

Multiplying through by x^2, we get:
-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2

As x \to 0, both -x^2 and x^2 approach 0. By the Squeeze Theorem:
\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0

These examples illustrate a variety of techniques used to evaluate limits, including direct substitution, factoring, L’Hôpital’s Rule, and the Squeeze Theorem.

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