Title: Limits
Course: Demonstrating STACK (v4.4.7)

You should use theorems on limits and the result that \[ \lim_{[value]\rightarrow 0}{\left(\frac{\sin([value])}[value]\right)}=1. \]
You should not use a Calculator or L'Hopital's rule.

Evaluate the following limit: \[ \lim_{[value]\rightarrow 0}{\left(\frac{\sin([value] [value])}{\sin([value][value])}\right)} .\]
The limit =
Feedback: We have \[ \lim_{[value]\rightarrow 0}{\left(\frac{\sin([value] [value])}{\sin([value][value])}\right)} = \lim_{[value]\rightarrow 0}{\left(\frac{\sin([value] [value])}{[value] [value]} \frac{[value] [value]}{\sin([value][value])} \frac{[value][value]}{[value][value]}\right)} = 1\cdot1\cdot\frac[value][value]= [value]\]

A \({\bf rational}\mbox{ } {\bf function}\) is a quotient of two polynomials. Evaluate the following limit of a rational function: \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] If the limit does not exist, enter the value 1111 as your answer.

The limit is:
Feedback: The limit of the function \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] is the same as the limit found by keeping only the highest power terms in the numerator and the denominator. Hence resulting in a limit of [value]. The limit of the function can be found by dividing the numerator and denominator by [value]. The function can thus be expressed as \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] As [value] tends to [value], all terms apart from the constants tend to zero. Hence the limit is [value].

A \({\bf rational}\mbox{ } {\bf function}\) is a quotient of two polynomials. Evaluate the following limit of a rational function: \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] If the limit does not exist, enter the value 1111 as your answer.

The limit is:
Feedback: The limit of the function \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] is the same as the limit found by keeping only the highest power terms in the numerator and the denominator. Hence the limit is undefined. The limit of the function can be found by dividing the numerator and denominator by [value]. The function can thus be expressed as \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] The limit of the numerator is [value] because all the other terms tend to zero. The limit of the denominator is zero. Hence the limit is undefined.

A \({\bf rational}\mbox{ } {\bf function}\) is a quotient of two polynomials. Evaluate the following limit of a rational function: \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] If the limit does not exist, enter the value 1111 as your answer.

The limit is:
Feedback: The limit of the function \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] is the same as the limit found by keeping only the highest power terms in the numerator and the denominator. Hence resulting in a limit of [value]. The limit of the function can be found by dividing the numerator and denominator by [value]. The function can thus be expressed as \[ \lim_{[value] \rightarrow [value]}{\left(\frac[value][value]\right)}. \] As [value] tends to [value], all terms apart from the constant tend to zero. Hence the limit is [value].

Evaluate the limit given below.

\[ \lim_{[value] \rightarrow [value]} {\left(\frac[value][value]\right) }.  \]

If the limit does not exist, enter the value 1111 as your answer.

The limit is:
Feedback: The limit of the function \[ \lim_{[value] \rightarrow [value]} {\left(\frac[value][value]\right) } \] is the same as the limit found by multiplying both numerator and denominator with the conjugate of the denominator: \[ \lim_{[value] \rightarrow [value]} {\left(\frac[value][value] \frac[value][value] \right) }, \] which equals to \[ \lim_{[value] \rightarrow [value]} {\left(\frac[value][value] \right) }, \] or, cancelling out the term in \( [value] \), equals \[ \lim_{[value] \rightarrow [value]} {\left( [value] \right) } = [value]. \]