Limits and Continuity

Using L'Hôpital's Rule immediately without any preliminary algebraic manipulation.

Direct substitution of $h=0$ into the expression.

Expanding $(\sqrt{a+h})^n$ using binomial expansion and then taking a limit.

Multiplying by the conjugate of the numerator to rationalize it before taking the limit.

The vertical asymptote becomes horizontal.

The function no longer has a vertical asymptote for any value of $b$.

It does not change; the vertical asymptote remains at $x=a$.

The vertical asymptote moves left if $b<3$ and right if $b>3$.

9/2.

The limit does not exist.

The limit is infinite.

15/4.

Evaluate limits from both sides of $x=5$ to check if they are equal.

Integrate $f(x)$ from $x=4$ to $x=6$ and divide by two.

Multiply $f(x)$ by $-\frac{1}{(5-x)^2}$ and evaluate at $x=5$.

Set $f(5)$ equal to zero since there's a discontinuity at $x=5$.

The derivative at point $c$, that is, $f'(c)$

Both one-sided limits $\lim_{x \to c^-} f(x)$ and $\lim_{x \to c^+} f(x)$

The limit as $x$ approaches $b$, i.e., $\lim_{x \to b} f(x)$

The definite integral from $a$ to $b$, $\int_{a}^{b} f(x) , dx$

The conjugate is (a+.

The conjugate is (a-.

The conjugate is (-a+.

The conjugate is (-a-.

The limit is larger than L since multiplying by c increases p(x).

The limit is smaller than L since multiplying by c decreases p(x).

The limit does not exist because changing p(x) changes its behavior as x approaches infinity.

The limit is still L since multiplying by a positive constant doesn't affect limits at infinity for root functions.

Analyzing a table and solving for the asymptote.

Graphing the function and using trigonometric identities.

Analyzing a table of values and using conjugates.

Graphing the function and analyzing a table of values.

$\int_{0}^{\frac{\pi}{4}}|\sin(x)-\cos(2x)|dx$

$\int_{0}^{\frac{\pi}{4}}(\sin(x)+\cos(2x))dx$

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}|\sin(x)-\cos(2x)|dx$

$\int_{0}^{\frac{\pi}{4}}(\sin(x)-\cos(2x))dx$

$\frac{L}{M}$ if M \neq 0

$0$

$LM$

$L+M$