Substitute x=0. This gives ( 0 − 0)/(03) = 0/0, an indeterminate form. Apply L'Hôpital's Rule.

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Here are the solutions for problems 9 through 15.

9. Evaluate: $\lim_{x \to 0} \frac{\sin x - x}{x^3}$

Substitute $x=0$ .
This gives $\frac{\sin 0 - 0}{0^3} = \frac{0}{0}$ , an indeterminate form. Apply L'Hôpital's Rule.
$\lim_{x \to 0} \frac{\sin x - x}{x^3} = \lim_{x \to 0} \frac{\frac{d}{dx}(\sin x - x)}{\frac{d}{dx}(x^3)} = \lim_{x \to 0} \frac{\cos x - 1}{3x^2}$
Substitute $x=0$ again.
This gives $\frac{\cos 0 - 1}{3(0)^2} = \frac{1 - 1}{0} = \frac{0}{0}$ , another indeterminate form. Apply L'Hôpital's Rule again.
$\lim_{x \to 0} \frac{\cos x - 1}{3x^2} = \lim_{x \to 0} \frac{\frac{d}{dx}(\cos x - 1)}{\frac{d}{dx}(3x^2)} = \lim_{x \to 0} \frac{-\sin x}{6x}$
Substitute $x=0$ again.
This gives $\frac{-\sin 0}{6(0)} = \frac{0}{0}$ , another indeterminate form. Apply L'Hôpital's Rule a third time.
$\lim_{x \to 0} \frac{-\sin x}{6x} = \lim_{x \to 0} \frac{\frac{d}{dx}(-\sin x)}{\frac{d}{dx}(6x)} = \lim_{x \to 0} \frac{-\cos x}{6}$
Substitute $x=0$ into the expression.
$\lim_{x \to 0} \frac{-\cos x}{6} = \frac{-\cos 0}{6} = \frac{-1}{6}$
The limit is $\boxed{-\frac{1}{6}}$ .

10. Evaluate: $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$
Substitute $x=0$ .
This gives $\frac{e^0 - 1 - 0}{0^2} = \frac{1 - 1 - 0}{0} = \frac{0}{0}$ , an indeterminate form. Apply L'Hôpital's Rule.
$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1 - x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{e^x - 1}{2x}$
Substitute $x=0$ again.
This gives $\frac{e^0 - 1}{2(0)} = \frac{1 - 1}{0} = \frac{0}{0}$ , another indeterminate form. Apply L'Hôpital's Rule again.
$\lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(2x)} = \lim_{x \to 0} \frac{e^x}{2}$
Substitute $x=0$ into the expression.
$\lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}$
The limit is $\boxed{\frac{1}{2}}$ .

11. Evaluate: $\lim_{x \to \infty} \frac{x^3}{\ln x}$
As $x \to \infty$ , $x^3 \to \infty$ and $\ln x \to \infty$ .
This is an indeterminate form $\frac{\infty}{\infty}$ . Apply L'Hôpital's Rule.
$\lim_{x \to \infty} \frac{x^3}{\ln x} = \lim_{x \to \infty} \frac{\frac{d}{dx}(x^3)}{\frac{d}{dx}(\ln x)} = \lim_{x \to \infty} \frac{3x^2}{\frac{1}{x}}$
Simplify the expression.
$\lim_{x \to \infty} \frac{3x^2}{\frac{1}{x}} = \lim_{x \to \infty} (3x^2 \cdot x) = \lim_{x \to \infty} 3x^3$
Evaluate the limit.
As $x \to \infty$ , $3x^3 \to \infty$ .
The limit is $\boxed{\infty}$ .

12. Evaluate: $\lim_{x \to 0^+} x \ln x$
As $x \to 0^+$ , $x \to 0$ and $\ln x \to -\infty$ .
This is an indeterminate form $0 \cdot (-\infty)$ . Rewrite the expression as a fraction.
$\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}}$
As $x \to 0^+$ , $\ln x \to -\infty$ and $\frac{1}{x} \to \infty$ .
This is an indeterminate form $\frac{-\infty}{\infty}$ . Apply L'Hôpital's Rule.
$\lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(\frac{1}{x})} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}}$
Simplify the expression.
$\lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} \left(\frac{1}{x} \cdot (-x^2)\right) = \lim_{x \to 0^+} (-x)$
Substitute $x=0$ into the expression.
$\lim_{x \to 0^+} (-x) = 0$
The limit is $\boxed{\text{0}}$ .

13. Evaluate: $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$
This is a standard definition of the mathematical constant $e$ .
$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$
The limit is $\boxed{e}$ .

14. Evaluate: $\lim_{x \to 0} \frac{\ln(1 + x) - x}{x^2}$
Substitute $x=0$ .
This gives $\frac{\ln(1 + 0) - 0}{0^2} = \frac{\ln 1 - 0}{0} = \frac{0}{0}$ , an indeterminate form. Apply L'Hôpital's Rule.
$\lim_{x \to 0} \frac{\ln(1 + x) - x}{x^2} = \lim_{x \to 0} \frac{\frac{d}{dx}(\ln(1 + x) - x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{\frac{1}{1 + x} - 1}{2x}$
Substitute $x=0$ again.
This gives $\frac{\frac{1}{1 + 0} - 1}{2(0)} = \frac{1 - 1}{0} = \frac{0}{0}$ , another indeterminate form. Apply L'Hôpital's Rule again.
$\lim_{x \to 0} \frac{\frac{1}{1 + x} - 1}{2x} = \lim_{x \to 0} \frac{\frac{d}{dx}((1 + x)^{-1} - 1)}{\frac{d}{dx}(2x)} = \lim_{x \to 0} \frac{-1(1 + x)^{-2}}{2}$
Simplify and substitute $x=0$ into the expression.
$\lim_{x \to 0} \frac{-\frac{1}{(1 + x)^2}}{2} = \frac{-\frac{1}{(1 + 0)^2}}{2} = \frac{-1}{2}$
The limit is $\boxed{-\frac{1}{2}}$ .

15. Evaluate: $\lim_{x \to \infty} x \sin \left(\frac{1}{x}\right)$
As $x \to \infty$ , $x \to \infty$ and $\sin\left(\frac{1}{x}\right) \to \sin(0) = 0$ .
This is an indeterminate form $\infty \cdot 0$ .
Let $y = \frac{1}{x}$ .
As $x \to \infty$ , $y \to 0$ . Substitute $x = \frac{1}{y}$ into the limit expression.
$\lim_{x \to \infty} x \sin \left(\frac{1}{x}\right) = \lim_{y \to 0} \frac{1}{y} \sin(y) = \lim_{y \to 0} \frac{\sin y}{y}$
This is a fundamental trigonometric limit.
$\lim_{y \to 0} \frac{\sin y}{y} = 1$
The limit is $\boxed{\text{1}}$ .

Substitute x=0. This gives ( 0 - 0)/(03) = (0)/(0), an indeterminate form. Apply L'Hôpital's Rule.

Step-by-step solution

Substitute $x=0$ .

Substitute $x=0$ again.

Substitute $x=0$ again.

Substitute $x=0$ into the expression.

Substitute $x=0$ .

Substitute $x=0$ again.

Substitute $x=0$ into the expression.

As $x \to \infty$ , $x^3 \to \infty$ and $\ln x \to \infty$ .

Simplify the expression.

Evaluate the limit.

As $x \to 0^+$ , $x \to 0$ and $\ln x \to -\infty$ .

As $x \to 0^+$ , $\ln x \to -\infty$ and $\frac{1}{x} \to \infty$ .

Simplify the expression.

Substitute $x=0$ into the expression.

This is a standard definition of the mathematical constant $e$ .

Substitute $x=0$ .

Substitute $x=0$ again.

Simplify and substitute $x=0$ into the expression.

As $x \to \infty$ , $x \to \infty$ and $\sin\left(\frac{1}{x}\right) \to \sin(0) = 0$ .

Let $y = \frac{1}{x}$ .

This is a fundamental trigonometric limit.

Still stuck on this one?

More Mathematics Questions

Substitute x=0. This gives ( 0 - 0)/(03) = (0)/(0), an indeterminate form. Apply L'Hôpital's Rule.

Step-by-step solution

Substitute x=0x=0x=0.

Substitute x=0x=0x=0 again.

Substitute x=0x=0x=0 again.

Substitute x=0x=0x=0 into the expression.

Substitute x=0x=0x=0.

Substitute x=0x=0x=0 again.

Substitute x=0x=0x=0 into the expression.

As x→∞x \to \inftyx→∞, x3→∞x^3 \to \inftyx3→∞ and ln⁡x→∞\ln x \to \inftylnx→∞.

Simplify the expression.

Evaluate the limit.

As x→0+x \to 0^+x→0+, x→0x \to 0x→0 and ln⁡x→−∞\ln x \to -\inftylnx→−∞.

As x→0+x \to 0^+x→0+, ln⁡x→−∞\ln x \to -\inftylnx→−∞ and 1x→∞\frac{1}{x} \to \inftyx1​→∞.

Simplify the expression.

Substitute x=0x=0x=0 into the expression.

This is a standard definition of the mathematical constant eee.

Substitute x=0x=0x=0.

Substitute x=0x=0x=0 again.

Simplify and substitute x=0x=0x=0 into the expression.

As x→∞x \to \inftyx→∞, x→∞x \to \inftyx→∞ and sin⁡(1x)→sin⁡(0)=0\sin\left(\frac{1}{x}\right) \to \sin(0) = 0sin(x1​)→sin(0)=0.

Let y=1xy = \frac{1}{x}y=x1​.

This is a fundamental trigonometric limit.

Still stuck on this one?

More Mathematics Questions

Substitute $x=0$ .

Substitute $x=0$ again.

Substitute $x=0$ again.

Substitute $x=0$ into the expression.

Substitute $x=0$ .

Substitute $x=0$ again.

Substitute $x=0$ into the expression.

As $x \to \infty$ , $x^3 \to \infty$ and $\ln x \to \infty$ .

As $x \to 0^+$ , $x \to 0$ and $\ln x \to -\infty$ .

As $x \to 0^+$ , $\ln x \to -\infty$ and $\frac{1}{x} \to \infty$ .

Substitute $x=0$ into the expression.

This is a standard definition of the mathematical constant $e$ .

Substitute $x=0$ .

Substitute $x=0$ again.

Simplify and substitute $x=0$ into the expression.

As $x \to \infty$ , $x \to \infty$ and $\sin\left(\frac{1}{x}\right) \to \sin(0) = 0$ .

Let $y = \frac{1}{x}$ .