Find Limits of Functions in Calculus

Note that we are looking for the limit as $x$ approaches 1 from the left ($x \to 1^-$ means $x$ approaches 1 by values smaller than 1). Hence:

$$x < 1$$ $$x - 1 < 0$$

If $x - 1 < 0$ then:

$$|x - 1| = -(x - 1)$$

Factor the numerator and simplify to obtain:

$$\lim_{x \to 1^-} \dfrac{(x - 1)(x + 3)}{-(x - 1)}$$ $$\lim_{x \to 1^-} \dfrac{(x - 1)(x + 3)}{-(x - 1)} = \lim_{x \to 1^-} -(x + 3) = -4$$

The final answer is:

$$\boxed{-4}$$

Although the limit in question is the ratio of two polynomials, $x = 5$ makes both the numerator and denominator equal to zero. We need to factor both the numerator and denominator as shown below.

$$\lim_{x \to 5} \dfrac{(x - 5)(x + 5)}{(x - 5)(x + 6)}$$

Simplify to obtain:

$$\lim_{x \to 5} \dfrac{x^2 - 25}{x^2 + x - 30} = \lim_{x \to 5} \dfrac{(x + 5)}{(x + 6)} = \dfrac{10}{11}$$

The final answer is:

$$\boxed{\dfrac{10}{11}}$$

We need to look at the limit from the left of 2 and the limit from the right of 2.

As $x$ approaches 2 from the left ($x \to 2^-$), $x - 2 < 0$, hence:

$$|x - 2| = -(x - 2)$$

Substitute to obtain the limit from the left of 2 as follows:

$$\lim_{x \to 2^-} \dfrac{x^2 + 4x - 12}{-(x - 2)}$$

Factor the numerator $x^2 + 4x - 12 = (x - 2)(x + 6)$:

$$= \lim_{x \to 2^-} \dfrac{(x - 2)(x + 6)}{-(x - 2)} = \lim_{x \to 2^-} -(x + 6) = -8$$

As $x$ approaches 2 from the right ($x \to 2^+$), $x - 2 > 0$, hence:

$$|x - 2| = x - 2$$

Substitute to obtain the limit from the right of 2 as follows:

$$\lim_{x \to 2^+} \dfrac{x^2 + 4x - 12}{x - 2} = \lim_{x \to 2^+} \dfrac{(x - 2)(x + 6)}{x - 2} = \lim_{x \to 2^+} (x + 6) = 8$$

The limit from the right of 2 (8) and the limit from the left of 2 (-8) are not equal, therefore the given two-sided limit DOES NOT EXIST.

As $x$ approaches -1 from the right, $\sqrt[3]{x+1}$ approaches 0 and $\ln(x+1)$ approaches $-\infty$, hence an indeterminate form $0 \times (-\infty)$.

$$\lim_{x \to -1^+} \sqrt[3]{x+1} \, \ln(x+1) = 0 \cdot (-\infty)$$

Let us rewrite the limit so that it is of the $\dfrac{\infty}{\infty}$ indeterminate form.

$$= \lim_{x \to -1^+} \dfrac{\ln(x+1)}{(x+1)^{-1/3}} = \dfrac{-\infty}{\infty}$$

We now use L'Hôpital's Rule and find the limit.

$$= \lim_{x \to -1^+} \dfrac{\dfrac{d}{dx}[\ln(x+1)]}{\dfrac{d}{dx}[(x+1)^{-1/3}]} = \lim_{x \to -1^+} \dfrac{\dfrac{1}{x+1}}{-\dfrac{1}{3}(x+1)^{-4/3}}$$ $$= \lim_{x \to -1^+} \dfrac{1}{x+1} \cdot \left( -\dfrac{3}{(x+1)^{-4/3}} \right) = \lim_{x \to -1^+} -3 (x+1)^{1/3} = 0$$

Thus, the limit is 0.

$$\boxed{0}$$

As $x$ gets larger, $x + 1$ gets larger, $\dfrac{1}{x+1}$ approaches zero, $e^{1/(x+1)}$ approaches 1, and $e^{1/(x+1)} - 1$ approaches 0; hence an indeterminate form: $\infty \times 0$.

$$\lim_{x \to \infty} - (x + 1)\left(e^{\dfrac{1}{x+1}} - 1\right) = -\infty \cdot 0$$

Let us rewrite the limit so that it is of the indeterminate form $\dfrac{0}{0}$:

$$= \lim_{x \to \infty} -\dfrac{e^{\dfrac{1}{x+1}} - 1}{\dfrac{1}{x+1}} = \dfrac{0}{0}$$

Apply L'Hôpital's theorem to find the limit:

$$\lim_{x \to \infty} -\dfrac{e^{\dfrac{1}{x+1}} - 1}{\dfrac{1}{x+1}} = -\lim_{x \to \infty} \dfrac{\dfrac{d}{dx}\left(e^{\dfrac{1}{x+1}} - 1\right)}{\dfrac{d}{dx}\left(\dfrac{1}{x+1}\right)}$$ $$= \lim_{x \to \infty} -\dfrac{\left(-\dfrac{1}{(x+1)^2}\right)e^{\dfrac{1}{x+1}}}{-\dfrac{1}{(x+1)^2}} = \lim_{x \to \infty} -e^{\dfrac{1}{x+1}} = -1$$

Hence:

$$\lim_{x \to \infty} - (x + 1)\left(e^{\dfrac{1}{x+1}} - 1\right) = -1$$

As $x$ approaches 9, both numerator and denominator approach 0. Multiply both numerator and denominator by the conjugate of the numerator:

$$\lim_{x \to 9} \dfrac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)}$$

Expand and simplify:

$$\lim_{x \to 9} \dfrac{x - 9}{(x - 9)(\sqrt{x} + 3)} = \lim_{x \to 9} \dfrac{1}{\sqrt{x} + 3}$$

Now find the limit by direct substitution:

$$\lim_{x \to 9} \dfrac{1}{\sqrt{x} + 3} = \dfrac{1}{6}$$

The range of the cosine function is:

$$-1 \le \cos x \le 1$$

Divide all terms of the above inequality by $x$, for $x > 0$:

$$-\dfrac{1}{x} \le \dfrac{\cos x}{x} \le \dfrac{1}{x}$$

Now as $x \to +\infty$, both $-\dfrac{1}{x}$ and $\dfrac{1}{x}$ approach 0. Hence, by the squeezing theorem, the limit is:

$$\lim_{x \to \infty} \dfrac{3 \cos x}{x} = 0$$

As $t$ approaches 0, both the numerator and denominator approach 0 and we have the $0 / 0$ indeterminate form.

$$\lim_{t \to 0} \dfrac{\sin t - t}{\tan t} = \dfrac{0}{0}$$

Hence, L'Hôpital's theorem is used to calculate the above limit as follows:

$$\lim_{t \to 0} \dfrac{\sin t - t}{\tan t} = \lim_{t \to 0} \dfrac{\dfrac{d}{dt}(\sin t - t)}{\dfrac{d}{dt}(\tan t)}$$ $$= \lim_{t \to 0} \dfrac{\cos t - 1}{\sec^2 t} = \dfrac{0}{1} = 0$$

We first factor out $16x^2$ under the square root of the denominator, take it out of the square root, and rewrite the limit as:

$$= \lim_{x \to \infty} \dfrac{3x}{4|x| \sqrt{1 + \dfrac{1}{16x^2}}}$$

Since $x$ approaches larger positive values (infinity), $|x| = x$ and $\dfrac{1}{16x^2}$ approaches 0. Simplify and find the limit:

$$= \lim_{x \to \infty} \dfrac{3x}{4x \sqrt{1 + \dfrac{1}{16x^2}}} = \dfrac{3}{4}$$

As $x$ approaches 2 from the left, then $x - 2$ approaches 0 from the left or $x - 2 < 0$. The numerator approaches 5 and the denominator approaches 0 from the left, hence the limit is given by:

$$\lim_{x \to 2^-} \dfrac{x^2 + 1}{x - 2} = -\infty$$

Factor $x^2$ in the denominator and simplify:

$$= \lim_{x \to \infty} \dfrac{3x^2}{x^2\left(4 + \dfrac{2}{x} - \dfrac{1}{x^2}\right)}$$

As $x$ takes large values (infinity), the terms $\dfrac{2}{x}$ and $\dfrac{1}{x^2}$ approach 0. Hence the limit is:

$$= \lim_{x \to \infty} \dfrac{3x^2}{4x^2} = \dfrac{3}{4}$$

Multiply numerator and denominator by $t$:

$$= \lim_{t \to 0} \dfrac{\sin(3t)}{t} \cdot \dfrac{t}{\sin(t)}$$

Rewrite the expression using algebraic manipulation:

$$= \lim_{t \to 0} \dfrac{\sin(3t)}{3t} \cdot \dfrac{3t}{\sin(t)} = \lim_{t \to 0} \dfrac{\sin(3t)}{3t} \cdot 3 \cdot \dfrac{t}{\sin(t)}$$

Use the known limit $\lim_{u \to 0} \dfrac{\sin u}{u} = 1$. Let $u = 3t$, so as $t \to 0$, $u \to 0$:

$$\lim_{t \to 0} \dfrac{\sin(3t)}{3t} = 1$$

Also:

$$\lim_{t \to 0} \dfrac{t}{\sin(t)} = 1$$

Hence, the limit is:

$$= 1 \cdot 3 \cdot 1 = 3$$

Factor $x^2$ inside the square root:

$$\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) = \lim_{x \to \infty} \left( \sqrt{x^2\left(1+\dfrac{1}{x} + \dfrac{1}{x^2}\right)} - x \right)$$

Use the fact that $\sqrt{x^2} = |x|$:

$$= \lim_{x \to \infty} \left( \sqrt{x^2} \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} - x \right) = \lim_{x \to \infty} \left( |x| \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} - x \right)$$

Since $x$ takes large positive values (infinity), then $|x| = x$ and both $\dfrac{1}{x}$ and $\dfrac{1}{x^2}$ approach 0. Hence we have the indeterminate form $\infty - \infty$:

$$= \lim_{x \to \infty} \left( x \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} - x \right) = \infty - \infty$$

Restart with the given limit again, multiply its numerator and denominator by the conjugate, and simplify:

$$= \lim_{x \to \infty} \dfrac{ \left( \sqrt{x^2 + x + 1} - x \right) \left( \sqrt{x^2 + x + 1} + x \right) }{ \sqrt{x^2 + x + 1} + x }$$ $$= \lim_{x \to \infty} \dfrac{ x^2 + x + 1 - x^2 }{ \sqrt{x^2 + x + 1} + x } = \lim_{x \to \infty} \dfrac{ x + 1 }{ \sqrt{x^2 + x + 1} + x }$$

Factor $x$ out of the numerator and denominator and simplify:

$$= \lim_{x \to \infty} \dfrac{ x \left( 1 + \dfrac{1}{x} \right) }{ x \left( \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} + 1 \right) } = \lim_{x \to \infty} \dfrac{ 1 + \dfrac{1}{x} }{ \sqrt{1 + \dfrac{1}{x} + \dfrac{1}{x^2}} + 1 }$$

As $x$ gets larger, the terms $\dfrac{1}{x}$ and $\dfrac{1}{x^2}$ approach zero and the limit evaluates to:

$$= \dfrac{1}{1 + 1} = \dfrac{1}{2}$$

Let $z = 1 / x$ so that as $x$ gets large, $z$ approaches 0. Substitute and calculate the limit as follows:

$$= \lim_{z \to 0} \dfrac{\sin z}{z} = 1$$

$$3$$

$$1$$

$$1$$

$$\dfrac{1}{4}$$

$$0$$

$$4$$