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- Calculus
- Limits
- Continuity
- Definition of the Derivative - First Principles
- Basic Differentiation Rules
- More Differentiation Rules
- Implicit Differentiation
- Higher Order Derivatives
- Curve Sketching
- First Derivative Test
- Second Derivative Test
- Derivatives of Trigonometric Functions
- Derivatives of Inverse Trigonometric Functions
- Derivatives of Logarithmic Functions
- Derivatives of Exponential Functions
- Antiderivative
- Indefinite Integral
- Applications of the Indefinite Integral
- The Definite Integral
- Area Under a Curve - Riemann Sums
- Area Under a Curve - Integration
- Area between Two Curves
- Volumes of Revolution
- Logarithmic Integrals
- Exponential Integrals
- Trigonometric Integrals
- Trigonometric Integrals of Other Forms
- More Integration Methods

Thinking of Limits as Approaching ...

We can think of limits as approaching something. Sometimes you can just plug in the value and work it out.

Other times we can't work something out directly, but we can see that it is getting closer and closer to something.

Let's consider the function \(f(x) = \frac{x^2-1}{x-1} \), what value is it approaching at \(x = 1\). If we plug \(x = 1\) into \(f(x) \), the value is undefined. However, if we evaluate the limit as it approaches \(x = 1\), we can see that the value is not undefined.

You can see that as *x* approaches 1, \(f(x)\) is approaching 2. But it is not true that when \(x=1\), \(f(x)=2\), so we state is as a limit,

\[ \lim_{x\to 1}\frac{x^2-1}{x-1} = 2\]

and we read it as "The limit of \(\frac{x^2-1}{x-1} \) as x approaches 1 is 2"

Suppose \(f\) is a real-valued function and \(a\) is a real number. We say that the limit of \(f(x)\) is \(L\) as \(x\) approaches \(a\) and write this as \[ \lim_{x\to a} f(x) = L \]

provided we can make \(f(x)\) as close to \(L\) as we want for all \(x\) sufficiently close to \(a\), from both sides, without actually letting \(x\) be \(a\).

Notice the definition states that limits are approached on both sides. So let's consider approaching \(f(x) = \frac{x^2+4x-12}{x^2-2x} \) as \(x\) approaches 2.

\[ \lim_{x\to 2} \frac{x^2+4x-12}{x^2-2x} \]

As we approached the function from x = 2.5 down to 2 we can see that it is \(f(x)\) is approaching 4. On the other side as x =1.5 up to 2, we see that it \(f(x)\) is also approaching 4. From the table, it appears that \(f(x)\) approaches 4 as x approaches 2, or

\[\lim_{x \to 2} \frac{x^2+4x-12}{x^2-2x} = 4 \]

The graphs shows the approach from both sides. There is an open circle around \(x=2\) because \(x=2\) does not exist (you cannot substitute \(x=2\) into the function.

- Last Updated: Aug 12, 2022 11:32 AM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=717032
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