Limits Practice

Because polynomials are continuous, substitute $x=4$ directly:

So the limit is $21$.

Factor the numerator:

For $x \ne 1$,

Now substitute $x=1$:

So the limit is $2$.

Approaching $2$ from the left means $x<2$, so the rule $f(x)=1$ applies.

Therefore,

This is a standard limit:

No. As $x$ gets close to $0$, the inside value $\frac{1}{x}$ grows without bound, so the sine function keeps oscillating between $-1$ and $1$.

Since it does not approach a single value, the limit does not exist.

The numerator and denominator have the same degree, so the limit is the ratio of leading coefficients:

So the limit is $\frac{5}{2}$.

As $x$ approaches $2$, the denominator $(x-2)^2$ approaches $0$ through positive values.

So the fraction grows without bound:

This means $x=2$ is a vertical asymptote.

Polynomials are continuous, so substitute $x=2$:

So the limit is $1$.

This is a standard limit:

Multiply by the conjugate:

Now substitute $x=9$:

So the limit is $\frac{1}{6}$.

Factor the numerator:

So for $x\ne 4$,

The limit as $x\to 4$ is

To make the function continuous, set

Find the one-sided limits.

From the left:

so

From the right:

Since the one-sided limits are different, the two-sided limit does not exist.

Factor the fraction first:

So the expression becomes

Now substitute $x=2$:

So the limit is $6$.

Rewrite the expression so the standard limit appears:

Then apply the standard limit:

So the limit is $3$.

Since

multiplying by $x^2 \ge 0$ gives

Both bounds go to $0$ as $x\to 0$, so by the squeeze theorem,

To find the horizontal asymptote, compute the limit as $x\to\infty$.

The numerator and denominator have the same degree, so the limit is the ratio of leading coefficients:

So the horizontal asymptote is

Rewrite the expression to match the standard limit:

Since

the limit is

Find the one-sided limits.

From the left:

From the right:

The one-sided limits are different, so no value of $k$ can make the function continuous at $x=1$.

As $x$ approaches $5$, the denominator $(x-5)^2$ approaches $0$ through positive values.

So the reading grows without bound:

This means $x=5$ is a vertical asymptote.

The numerator and denominator have the same degree, so the limit is the ratio of leading coefficients:

So the model approaches $\frac{7}{2}$ in the long run.

The expression has the indeterminate form $\frac{0}{0}$, so L'Hopital's rule applies.

Differentiate the numerator and denominator:

Now substitute $x=0$:

So the limit is $2$.

Compute the one-sided limits.

From the left:

From the right:

The one-sided limits are different, so $F$ is not continuous at $w=5$. This is a jump discontinuity.

Factor the numerator:

So for $x\ne 3$,

The limit as $x\to 3$ is

Since the limit exists but the original function is undefined at $x=3$, the discontinuity is removable.

To remove it, define $g(3)=6$.

Start with the expression:

To make this less than $\varepsilon$, it is enough to require

So choose

Then whenever $0<|x-2|<\delta$,

which proves the limit.

First find the limit from the left:

So the value needed for continuity at $x=2$ is

That gives the equation

The condition $f(3)=10$ gives

Subtract the first equation from the second:

Then

so

Use the squeeze theorem first on $x\sin(1/x)$:

So

Also,

Since the denominator approaches a nonzero number, the quotient approaches

So the limit is $0$.

Factor the numerator:

So for $x\ne 1$,

The limit as $x\to 1$ is

So continuity at $x=1$ requires

For end behavior, the function behaves like $x+1$ as $x\to\infty$, so it grows without bound.

Therefore, $f$ does not have a horizontal asymptote.