Question
(1) Evaluate the following limits. State the asymptotes, if any:
(i)
(ii)
(i)
(ii)
Ask by Reid Schneider. in Canada
Jan 26,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
Limits and Asymptotes:
-
Limit (i):Asymptote: Vertical asymptote at
. -
Limit (ii):Asymptote: Horizontal asymptote at
.
Solution

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Mind Expander
Let’s evaluate the limits step by step, starting with the first one.
(i) To find the limit
, we first substitute
into the denominator:
This gives us a
in the denominator. We then need to factor the denominator. The polynomial
can be factored as follows:
Thus, the limit can be rewritten as:
We can now cancel
in the numerator and denominator (noting that we are approaching
, so we are not actually at 3):
Substituting
:
Thus, the limit is
and we note that there is a vertical asymptote at
since the denominator equals zero there.
(ii) Now we evaluate
.
To simplify for large negative
, we factor out
from the square roots in both the numerator and denominator.
In the numerator:
Thus, we rewrite the numerator:
In the denominator:
Thus, the denominator becomes:
Now we can simplify the limit:
[
\lim_{x \rightarrow -\infty} \frac{x(1 + \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}})}{2x(1 + \sqrt{1 + \frac{5}{4x} - \frac{1}{4x^2}})} = \lim_{x \rightarrow -\infty} \frac{1 + \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}}{2(1 + \sqrt{1 + \frac{5}{4x} - \
\lim_{x \rightarrow -\infty} \frac{x(1 + \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}})}{2x(1 + \sqrt{1 + \frac{5}{4x} - \frac{1}{4x^2}})} = \lim_{x \rightarrow -\infty} \frac{1 + \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}}{2(1 + \sqrt{1 + \frac{5}{4x} - \