Math  /  Calculus

QuestionThe First Gate Upon arriving at the first gate, you ask your computer to translate the riddle. "These appear to be limits", your computer replies. "If you provide me with the answer to each of below questions, I will compile, translate, and communicate the passcode to the gate. There is note here that the parameter aa represents a real number such that 0<a<90<a<9." Your computer also reminds you to type "infinity" for \infty, "-infinity" for -\infty, and "NA" if the limit does not exist. You grab your notebook and pencil and begin. limxax+9xa=limxa+x+9xa=limxax+9xa=\begin{array}{r} \lim _{x \rightarrow a^{-}} \frac{x+9}{x-a}= \\ \lim _{x \rightarrow a^{+}} \frac{x+9}{x-a}= \\ \lim _{x \rightarrow a} \frac{x+9}{x-a}= \end{array} \square \square \square

Studdy Solution
Determine the two-sided limit limxax+9xa\lim_{x \rightarrow a} \frac{x+9}{x-a}.
Since the left-hand limit and the right-hand limit are not equal (-\infty \neq \infty), the two-sided limit does not exist.
Therefore, the limit is:
limxax+9xa=NA\lim_{x \rightarrow a} \frac{x+9}{x-a} = \text{NA}
The values of the limits are:
limxax+9xa=limxa+x+9xa=limxax+9xa=NA\begin{array}{r} \lim _{x \rightarrow a^{-}} \frac{x+9}{x-a}= -\infty \\ \lim _{x \rightarrow a^{+}} \frac{x+9}{x-a}= \infty \\ \lim _{x \rightarrow a} \frac{x+9}{x-a}= \text{NA} \end{array}

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord