Problem 19
Question
In Exercises 9-36, find the limit (if it exists). Use a graphing utility to verify your result graphically. $$\lim_{x \to 1} \dfrac{x^2+x-2}{x^2-3x+2}$$
Step-by-Step Solution
Verified Answer
The limit of the given expression as \(x\) approaches 1 is \(-3\)
1Step 1: Simplify
First, factorise both the numerator and the denominator. This would make both of them easier to deal with. The expression \(x^2 + x - 2\) can be expressed as \((x-1)(x+2)\) and \(x^2 - 3x + 2\) can be expressed as \((x-1)(x-2)\). So, the expression will become: \(\lim_{x \to 1} \dfrac{(x-1)(x+2)}{(x-1)(x-2)}\)
2Step 2: Compute the Limit
Next, cancel out the common factor \((x-1)\) in the numerator and denominator, simplifying the expression to \(\lim_{x \to 1} \dfrac{x+2}{x-2}\). Now, plug in \(x=1\) into the expression to find the solution to the limit: \((1+2)/(1-2)\), which equals \(-3\).
3Step 3: Verify Graphically
To verify the finding graphically, plot the function \(\dfrac{x+2}{x-2}\) with \(x\)‘s range close to 1, and observe the function's behavior as it approaches 1 from both sides. The value should approximate \(-3\), which matches the computed limit.
Key Concepts
Factoring PolynomialsRational FunctionsGraphical Verification
Factoring Polynomials
Factoring polynomials is a fundamental skill in algebra that involves breaking down a complex polynomial expression into products of simpler polynomials. In our given problem, we are dealing with the expression \(\lim_{x \to 1} \dfrac{x^2+x-2}{x^2-3x+2}\). To simplify this, start by factoring the numerator and the denominator separately.
For the numerator \(x^2 + x - 2\), identify two numbers that multiply to -2 and add to 1. These numbers are 2 and -1, yielding the factorization \((x-1)(x+2)\).
For the denominator \(x^2 - 3x + 2\), look for two numbers that multiply to 2 and add to -3. These are -1 and -2, giving the factorization \((x-1)(x-2)\). After factorization, the original expression becomes:
\[\frac{(x-1)(x+2)}{(x-1)(x-2)}\]
Notice that \((x-1)\) is a common factor in both the numerator and the denominator, allowing it to be canceled out, simplifying the expression. This is a critical step for calculating limits, as it often removes indeterminate forms like \(\frac{0}{0}\), making it possible to directly substitute the value of \(x\).
For the numerator \(x^2 + x - 2\), identify two numbers that multiply to -2 and add to 1. These numbers are 2 and -1, yielding the factorization \((x-1)(x+2)\).
For the denominator \(x^2 - 3x + 2\), look for two numbers that multiply to 2 and add to -3. These are -1 and -2, giving the factorization \((x-1)(x-2)\). After factorization, the original expression becomes:
\[\frac{(x-1)(x+2)}{(x-1)(x-2)}\]
Notice that \((x-1)\) is a common factor in both the numerator and the denominator, allowing it to be canceled out, simplifying the expression. This is a critical step for calculating limits, as it often removes indeterminate forms like \(\frac{0}{0}\), making it possible to directly substitute the value of \(x\).
Rational Functions
Rational functions are fractions where the numerator and the denominator are both polynomials. In our example, substituting the polynomials into a rational function:
\[ \frac{x^2+x-2}{x^2-3x+2} \]
becomes \( \frac{(x-1)(x+2)}{(x-1)(x-2)} \) after factoring. Simplifying this by canceling the common terms, we are left with \( \frac{x+2}{x-2} \).
Rational functions can often lead to undefined points where the denominator is zero. In this exercise, \((x-1)\) was such a factor, which we effectively removed to avoid division by zero at \(x=1\). After removing the factor, the limit becomes easier to compute.
When the limit approaches a value where the original function was undefined, it shows the behavior of the function very close to that point, providing insight into potential asymptotic behaviors or holes in the graph. This is why rational functions are particularly important in calculus and require careful analysis through factoring and simplification.
\[ \frac{x^2+x-2}{x^2-3x+2} \]
becomes \( \frac{(x-1)(x+2)}{(x-1)(x-2)} \) after factoring. Simplifying this by canceling the common terms, we are left with \( \frac{x+2}{x-2} \).
Rational functions can often lead to undefined points where the denominator is zero. In this exercise, \((x-1)\) was such a factor, which we effectively removed to avoid division by zero at \(x=1\). After removing the factor, the limit becomes easier to compute.
When the limit approaches a value where the original function was undefined, it shows the behavior of the function very close to that point, providing insight into potential asymptotic behaviors or holes in the graph. This is why rational functions are particularly important in calculus and require careful analysis through factoring and simplification.
Graphical Verification
Graphical verification is a vital step to confirm your analytical results visually by using a graphing utility or sketching the function's graph. For this problem, plotting \( \frac{x+2}{x-2} \) with a focus around \(x=1\) helps to visualize the behavior of the function near the point of interest.
When you create the graph:
When you create the graph:
- Pay attention to the approach towards \(x=1\) from both sides.
- Notice any sharp changes or vertical asymptotes, which indicate where the function's value may be approaching a specific number, confirming our limit calculation.
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