Question

$$\lim _{x \rightarrow 1^{+}} \frac{x^{2}-5 x+6}{x-1}$$

Short answer

Answer: The limit as x approaches 1 from the positive side of the given expression is -3.

Step-by-step solution

Try to simplify the expression

First try to simplify the expression by factoring the numerator: $$\frac{x^2-5x+6}{x-1}=\frac{(x-2)(x-3)}{x-1}$$

Look for cancellations or simplifications

In this expression, there are no common factors in the numerator and denominator, so we cannot simplify it further.

Try direct substitution of x=1

Directly substitute x=1 and check if it results in an indeterminate form (0/0 or ∞/∞): $$\frac{(1-2)(1-3)}{1-1}=\frac{-1\times -2}{0}$$

Identify the indeterminate form

As we see, substituting x=1 results in a division by zero, which means the expression is undefined at x=1. So we cannot use direct substitution to find the limit.

Find the limit using derivative (L'Hôpital's Rule)

Since direct substitution doesn't work, we will use L'Hôpital's Rule to find the limit. This rule states that if the limit of the ratio of the derivatives of the numerator and denominator exists, then it is equal to the limit of the original problem. So we need to differentiate the numerator and denominator and then find the limit of their ratio. Differentiating the numerator with respect to x: $$(x^2-5x+6)'=(x-2)(x-3)'(x+2)(x+1)'=2x-5$$ Differentiating the denominator with respect to x: $$(x-1)'=1$$ Now, we can find the limit of the ratio of derivatives as x approaches 1: $$\lim_{x \to 1^+} \frac{(x^2-5x+6)'}{(x-1)'}=\lim_{x \to 1^+} \frac{2x-5}{1}=\frac{2(1)-5}{1}=\frac{-3}{1}=-3$$

Final Answer

The limit as x approaches 1 from the positive side of the given expression is -3: $$\lim _{x \rightarrow 1^{+}} \frac{x^{2}-5 x+6}{x-1}=-3$$

Key concepts

Understanding Limits

The concept of a limit is integral in calculus, helping us understand how a function behaves as its input approaches a certain value. Consider the expression \( \lim _{x \to 1^+} \frac{x^2-5x+6}{x-1} \). Here, we want to know what happens to \( \frac{x^2-5x+6}{x-1} \) as \( x \) gets closer to 1 from the right side. This involves evaluating the function's behavior rather than its exact value at that point.

If you attempt plugging in \( x = 1 \) directly, you'll get \( \frac{0}{0} \), an undefined form. Instead, you've observed a pattern or trend. By using a systematic approach, like factoring or derivative techniques—such as L'Hôpital's Rule, which acts as a powerful tool in evaluating such limits—you can often resolve this ambiguity.

In simple terms, the limit tells us the general tendency of the function's output, offering insights where direct evaluation hits a wall.

Indeterminate Forms: Not Always Undefined

Indeterminate forms like \( \frac{0}{0} \) aren't nonsense, even though they seem undefined initially. They're a prompt to dig deeper. An expression like \( \lim_{x \to 1} \frac{x^2-5x+6}{x-1} = \frac{-1 \times -2}{0} \) is a typical indeterminate form.

Mathematically, these forms suggest a restructuring of the given function. Indeterminate doesn't mean unsolvable; it means that with proper application of calculus methods, such as simplification or L'Hôpital's Rule, the limit can indeed be determined. These forms encourage exploring the dynamic nature of functions, leading to a clearer understanding of the value or trend at a particular point.

• Think of indeterminate forms as doors—which need specific keys (like derivatives or simplification) to unlock insight.
• Rather than yielding a result immediately, they point towards more significant exploratory steps.

Derivatives: The Tool for L'Hôpital's Rule

Derivatives play a crucial role when dealing with limits, especially when applying L'Hôpital's Rule. A derivative represents how fast a function changes at any given point, accessing the instantaneous rate of change. This gives us a new perspective to analyze the problem.

In L'Hôpital's Rule, if you face \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), differentiate both the numerator and denominator separately. For our problem, the original expression \( \lim _{x \to 1^+} \frac{(x^2-5x+6)'}{(x-1)'} \) gets replaced by \( \frac{2x-5}{1} \) upon differentiation.

The limit of \( \frac{2x-5}{1} \) as \( x \) approaches 1 becomes straightforward to calculate, ultimately giving \( -3 \). Hence, derivatives simplify the evaluation process by transforming an indeterminate form into a straightforward calculation.

By understanding derivatives, not only do you solve complex limits, but you also gain deeper insight into the function's behavior.