In order to evaluate limits, we can take a look at their various properties to help simplify the process. In this instance, assume that both \(f\) and \(g\) have limits that exist at \(x = a\) and that \(c\) represents the constant.
The properties can be summarized as such:
| Constant | \(\lim\limits_{x\to a}c=c\) |
| Constant times a Function | \(\lim\limits_{x\to a}[c \cdot f(x)]= c\lim\limits_{x\to a}f(x)\) |
| Sum of Functions | \(\lim\limits_{x\to a}[f(x)+g(x)] = \lim\limits_{x\to a}f(x)+\lim\limits_{x\to a}g(x)\) |
| Difference of Functions | \(\lim\limits_{x\to a}[f(x)-g(x)] = \lim\limits_{x\to a}f(x)-\lim\limits_{x\to a}g(x)\) |
| Product of Functions | \(\lim\limits_{x\to a}[f(x)\cdot g(x)] = \lim\limits_{x\to a}f(x)\cdot\lim\limits_{x\to a}g(x)\) |
| Quotient of Functions | \(\lim\limits_{x\to a} \left[\cfrac{f(x)}{g(x)}\right] = \cfrac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)}\) |
| Function Raised to an Exponent | \(\lim\limits_{x\to a}[f(x)]^n=\lim\limits_{x\to a}[f(x)]^n\) |
| nth root of a Function, 0 < n | \(\lim\limits_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{\lim\limits_{x\to a}f(x)}\) |
| Polynomial Function | \(\lim\limits_{x\to a}p(x)=p(a)\) |
Here are some additional pointers for helping determine the limits for each expression:
Evaluate the following limits:
i. First, as the entire function is placed under a square root, we can use the Function Raised to Exponent Property to determine that its being raised to a power of \( \frac{1}{2} \).
Next, we can use the Quotient Property to find the limits of the respective pieces:
\(= \sqrt{\cfrac{\lim\limits_{x\to5}x^2}{\lim\limits_{x\to5}x-1}}\)
Then, we can use the Function Raised to an Exponent Property in the numerator to substitute the limit for \(x\):
\(= \sqrt{\cfrac{(\lim\limits_{x\to5}x)^2}{\lim\limits_{x\to5}x-1}}\)
Finally, we can use the Constant Property to substitute the limit value (in this case, \(5\)) for \(x\). After, we can simplify to get the final result:
\(= \sqrt{\cfrac{(5)^2}{5-1}}\)
\(= \sqrt{\cfrac{25}{4}}\)
\(= \cfrac{5}{2}\)
Therefore, we can determine that this limit DOES exists.
ii. First, we can test the limit from its left side:
We can subsitute a value close to the limit (in this case \(2.9\) for \(x\). We can then simplify the function to evaluate the limit:
\(= \sqrt{9-(2.9)^2}\)
\(= \sqrt{9-8.41}\)
\(= \sqrt{0.59}\)
\(= 0.77\)
We can determine that testing the limit from its left side provides us with a real value.
Next, we can evaluate the limit from its right side:
\(= \lim\limits_{x\to3^{+}}\sqrt{9-x^2}\)
We can substitute this value for \(x\) in the function. We can then simplify it to evaluate the limit:
\(= \sqrt{9-(3.1)^2}\)
\(= \sqrt{9-9.61}\)
\(= \sqrt{-0.61}\)
\(= \text{UNDEFINED}\)
When we compare the limits on both sides, we get the following result:
\(\lim\limits_{x\to3^{-}} \neq \lim\limits_{x\to3^{+}}\)
Therefore, we can determine that this limit DOES NOT exist as it only represents a one-sided limit. Since the domain is restricted, we can determine that the function doesn't exist on the left side.
\(\lim\limits_{x\to2}\cfrac{|2x-4|}{x-2}\)
When we evaluate the limit initially, we get the following result:
\(= \cfrac{|2(2)-4|}{2-2}\)
\(= \cfrac{|4-4|}{0}\)
\(= \cfrac{0}{0}\)
Since this limit evaluates to \(\cfrac{0}{0}\), we can take out a common factor:
\(\cfrac{2|x-2|}{x-2}\)
Since the numerator represents an Absolute Function, we have to test the limit on both sides separately:
We will start with evaluating \(\lim\limits_{x\to 2^{+}}\):
\(= \lim\limits_{x\to 2^{+}}\cfrac{2\cancel{(x-2)}}{\cancel{x-2}}\)
\(= \lim\limits_{x\to2^{+}} 2\)
\(= 2\)
Next, we will evaluate \(\lim\limits_{x\to2^{-}}\). Since this is similar to evaluating a piecewise function, we will add a negative sign to the numerator:
\(= \lim\limits_{x\to2^{-}}\cfrac{-2\cancel{(x-2)}}{\cancel{x-2}}\)
\(= \lim\limits_{x\to2^{-}} -2\)
\(= -2\)
Finally, we can compare the values on both sides of the limit:
\(2 \neq -2\) OR \(\lim\limits_{x\to2^{+}} \neq \lim\limits_{x\to2^{-}}\)
Therefore, we can determine that this limit DOES NOT exist since the value on both sides of the limit aren't the same.
\(\lim\limits_{x\to∞}\cfrac{4 - 3x^3 + 2x}{8 - x^2 + 7x^3}\)
Since the limit is ∞, we can evaluate this function by dividing it by its highest power (in this case, \(3\)):
\(= \lim\limits_{x\to∞}\cfrac{\cfrac{4}{x^3} - \cfrac{3x^3}{x^3} + \cfrac{2x}{x^3}}{\cfrac{8}{x^3} - \cfrac{x^2}{x^3} + \cfrac{7x^3}{x^3}}\)
We can now simplify this function to get our final result:
\(= \lim\limits_{x\to ∞}\cfrac{\cancel{\cfrac{4}{x^3}} - \cfrac{3x^3}{x^3} + \cancel{\cfrac{2x}{x^3}}}{\cancel{\cfrac{8}{x^3}} - \cancel{\cfrac{x^2}{x^3}} + \cfrac{7x^3}{x^3}}\)
\(= \cfrac{\cfrac{-3\cancel{x^3}}{\cancel{x^3}}}{\cfrac{7\cancel{x^3}}{\cancel{x^3}}}\)
\(= \cfrac{-3}{7}\)
Therefore, we can determine that this limit DOES exists since it evaluates to a regular value.
\(\lim\limits_{x\to0}\cfrac{\sqrt{4+x}-2}{x}\)
When we immediately evaluate the limit, we get the following result:
\(= \lim\limits_{x\to0}\cfrac{\sqrt{4+0}-2}{0}\)
\(= \lim\limits_{x\to0}\cfrac{2-2}{0}\)
\(= \cfrac{0}{0}\)
Since this results in \(\cfrac{0}{0}\), we can rationalize by multiplying this function by its conjugate. This will eliminate the square root in the numerator and ensure that the denominator won't be \(0\):
\(= \lim\limits_{x\to0} \left(\cfrac{\sqrt{4+x}-2}{x}\right) \left(\cfrac{\sqrt{4+x}+2}{\sqrt{4+x}+2}\right)\)
\(= \lim\limits_{x\to0}\cfrac{4 + x - 4}{x(\sqrt{4+x}+2)}\)
Next, we can simplify this function:
\(= \lim\limits_{x\to0}\cfrac{\cancel{x}}{\cancel{x}(\sqrt{4+x}+2)}\)
\(= \lim\limits_{x\to0}\cfrac{1}{\sqrt{4+x}+2}\)
Finally, we can evaluate the limit to determine if it exists:
\(= \cfrac{1}{\sqrt{4+0}+2}\)
\(= \cfrac{1}{2+2}\)
\(= \cfrac{1}{4}\)
Therefore, we can determine that this limit DOES exist.