Evaluate the following limit problems: lim (sin x)/x as x approaches 0 lim (1-cos x)/(x2) as x approaches 0…

Question

Evaluate the following limit problems: lim (sin x)/x as x approaches 0 lim (1-cos x)/(x2) as x approaches 0 lim (ln x)/x as x approaches infinity lim x/(e^x) as x approaches infinity

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here are the solutions for these limit problems.

1. Evaluate: $\lim_{x \to 0} \frac{\sin x}{x}$

This is a fundamental trigonometric limit.
$\lim_{x \to 0} \frac{\sin x}{x} = 1$
The limit is $\boxed{\text{1}}$ .

2. Evaluate: $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$
This is a fundamental trigonometric limit.
$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
The limit is $\boxed{\frac{1}{2}}$ .

3. Evaluate: $\lim_{x \to \infty} \frac{\ln x}{x}$
As $x \to \infty$ , both $\ln x \to \infty$ and $x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .
Apply L'Hôpital's Rule.
$\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x)}$
Differentiate the numerator and the denominator.
$\lim_{x \to \infty} \frac{\frac{1}{x}}{1} = \lim_{x \to \infty} \frac{1}{x}$
Evaluate the limit.
$\lim_{x \to \infty} \frac{1}{x} = 0$
The limit is $\boxed{\text{0}}$ .

4. Evaluate: $\lim_{x \to \infty} \frac{x}{e^x}$
As $x \to \infty$ , both $x \to \infty$ and $e^x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .
Apply L'Hôpital's Rule.
$\lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{\frac{d}{dx}(x)}{\frac{d}{dx}(e^x)}$
Differentiate the numerator and the denominator.
$\lim_{x \to \infty} \frac{1}{e^x}$
Evaluate the limit.
As $x \to \infty$ , $e^x \to \infty$ , so $\frac{1}{e^x} \to 0$ .
$\lim_{x \to \infty} \frac{1}{e^x} = 0$
The limit is $\boxed{\text{0}}$ .

5. Evaluate: $\lim_{x \to 0} \frac{e^x - 1}{x}$
This is a fundamental limit related to the exponential function.
$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
The limit is $\boxed{\text{1}}$ .

6. Evaluate: $\lim_{x \to 1} \frac{\ln x}{x - 1}$
As $x \to 1$ , $\ln x \to \ln 1 = 0$ and $x - 1 \to 1 - 1 = 0$ , which is an indeterminate form $\frac{0}{0}$ .
Apply L'Hôpital's Rule.
$\lim_{x \to 1} \frac{\ln x}{x - 1} = \lim_{x \to 1} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x - 1)}$
Differentiate the numerator and the denominator.
$\lim_{x \to 1} \frac{\frac{1}{x}}{1} = \lim_{x \to 1} \frac{1}{x}$
Substitute $x=1$ into the expression.
$\lim_{x \to 1} \frac{1}{x} = \frac{1}{1} = 1$
The limit is $\boxed{\text{1}}$ .

7. Evaluate: $\lim_{x \to 0} \frac{\tan x - x}{x^3}$
As $x \to 0$ , $\tan x - x \to 0 - 0 = 0$ and $x^3 \to 0$ , which is an indeterminate form $\frac{0}{0}$ .
Apply L'Hôpital's Rule.
$\lim_{x \to 0} \frac{\tan x - x}{x^3} = \lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2}$
This is still an indeterminate form $\frac{0}{0}$ since $\sec^2 0 - 1 = 1 - 1 = 0$ .
Apply L'Hôpital's Rule again.
$\lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2} = \lim_{x \to 0} \frac{2 \sec x (\sec x \tan x)}{6x} = \lim_{x \to 0} \frac{2 \sec^2 x \tan x}{6x} = \lim_{x \to 0} \frac{\sec^2 x \tan x}{3x}$
This is still an indeterminate form $\frac{0}{0}$ .
Apply L'Hôpital's Rule a third time.
$\lim_{x \to 0} \frac{\sec^2 x \tan x}{3x} = \lim_{x \to 0} \frac{(2 \sec x (\sec x \tan x)) \tan x + \sec^2 x (\sec^2 x)}{3}$ $= \lim_{x \to 0} \frac{2 \sec^2 x \tan^2 x + \sec^4 x}{3}$
Substitute $x=0$ .
$\frac{2 \sec^2 0 \tan^2 0 + \sec^4 0}{3} = \frac{2(1)^2(0)^2 + (1)^4}{3} = \frac{0 + 1}{3} = \frac{1}{3}$
The limit is $\boxed{\frac{1}{3}}$ .

8. Evaluate: $\lim_{x \to \infty} \frac{x^2 + 1}{e^x}$
As $x \to \infty$ , both $x^2 + 1 \to \infty$ and $e^x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .
Apply L'Hôpital's Rule.
$\lim_{x \to \infty} \frac{x^2 + 1}{e^x} = \lim_{x \to \infty} \frac{\frac{d}{dx}(x^2 + 1)}{\frac{d}{dx}(e^x)} = \lim_{x \to \infty} \frac{2x}{e^x}$
This is still an indeterminate form $\frac{\infty}{\infty}$ .
Apply L'Hôpital's Rule again.
$\lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{\frac{d}{dx}(2x)}{\frac{d}{dx}(e^x)} = \lim_{x \to \infty} \frac{2}{e^x}$
Evaluate the limit.
As $x \to \infty$ , $e^x \to \infty$ , so $\frac{2}{e^x} \to 0$ .
$\lim_{x \to \infty} \frac{2}{e^x} = 0$
The limit is $\boxed{\text{0}}$ .

Evaluate the following limit problems: lim (sin x)/x as x approaches 0 lim (1-cos x)/(x2) as x approaches 0 lim (ln x)/x as x approaches infinity lim x/(e^x) as x approaches infinity

Step-by-step solution

This is a fundamental trigonometric limit.

This is a fundamental trigonometric limit.

As $x \to \infty$ , both $\ln x \to \infty$ and $x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .

Differentiate the numerator and the denominator.

Evaluate the limit.

As $x \to \infty$ , both $x \to \infty$ and $e^x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .

Differentiate the numerator and the denominator.

Evaluate the limit.

This is a fundamental limit related to the exponential function.

As $x \to 1$ , $\ln x \to \ln 1 = 0$ and $x - 1 \to 1 - 1 = 0$ , which is an indeterminate form $\frac{0}{0}$ .

Differentiate the numerator and the denominator.

Substitute $x=1$ into the expression.

As $x \to 0$ , $\tan x - x \to 0 - 0 = 0$ and $x^3 \to 0$ , which is an indeterminate form $\frac{0}{0}$ .

This is still an indeterminate form $\frac{0}{0}$ since $\sec^2 0 - 1 = 1 - 1 = 0$ .

This is still an indeterminate form $\frac{0}{0}$ .

Substitute $x=0$ .

As $x \to \infty$ , both $x^2 + 1 \to \infty$ and $e^x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .

This is still an indeterminate form $\frac{\infty}{\infty}$ .

Evaluate the limit.

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More Mathematics Questions

Evaluate the following limit problems: lim (sin x)/x as x approaches 0 lim (1-cos x)/(x2) as x approaches 0 lim (ln x)/x as x approaches infinity lim x/(e^x) as x approaches infinity

Step-by-step solution

This is a fundamental trigonometric limit.

This is a fundamental trigonometric limit.

As x→∞x \to \inftyx→∞, both ln⁡x→∞\ln x \to \inftylnx→∞ and x→∞x \to \inftyx→∞, which is an indeterminate form ∞∞\frac{\infty}{\infty}∞∞​.

Differentiate the numerator and the denominator.

Evaluate the limit.

As x→∞x \to \inftyx→∞, both x→∞x \to \inftyx→∞ and ex→∞e^x \to \inftyex→∞, which is an indeterminate form ∞∞\frac{\infty}{\infty}∞∞​.

Differentiate the numerator and the denominator.

Evaluate the limit.

This is a fundamental limit related to the exponential function.

As x→1x \to 1x→1, ln⁡x→ln⁡1=0\ln x \to \ln 1 = 0lnx→ln1=0 and x−1→1−1=0x - 1 \to 1 - 1 = 0x−1→1−1=0, which is an indeterminate form 00\frac{0}{0}00​.

Differentiate the numerator and the denominator.

Substitute x=1x=1x=1 into the expression.

As x→0x \to 0x→0, tan⁡x−x→0−0=0\tan x - x \to 0 - 0 = 0tanx−x→0−0=0 and x3→0x^3 \to 0x3→0, which is an indeterminate form 00\frac{0}{0}00​.

This is still an indeterminate form 00\frac{0}{0}00​ since sec⁡20−1=1−1=0\sec^2 0 - 1 = 1 - 1 = 0sec20−1=1−1=0.

This is still an indeterminate form 00\frac{0}{0}00​.

Substitute x=0x=0x=0.

As x→∞x \to \inftyx→∞, both x2+1→∞x^2 + 1 \to \inftyx2+1→∞ and ex→∞e^x \to \inftyex→∞, which is an indeterminate form ∞∞\frac{\infty}{\infty}∞∞​.

This is still an indeterminate form ∞∞\frac{\infty}{\infty}∞∞​.

Evaluate the limit.

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More Mathematics Questions

As $x \to \infty$ , both $\ln x \to \infty$ and $x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .

As $x \to \infty$ , both $x \to \infty$ and $e^x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .

As $x \to 1$ , $\ln x \to \ln 1 = 0$ and $x - 1 \to 1 - 1 = 0$ , which is an indeterminate form $\frac{0}{0}$ .

Substitute $x=1$ into the expression.

As $x \to 0$ , $\tan x - x \to 0 - 0 = 0$ and $x^3 \to 0$ , which is an indeterminate form $\frac{0}{0}$ .

This is still an indeterminate form $\frac{0}{0}$ since $\sec^2 0 - 1 = 1 - 1 = 0$ .

This is still an indeterminate form $\frac{0}{0}$ .

Substitute $x=0$ .

As $x \to \infty$ , both $x^2 + 1 \to \infty$ and $e^x \to \infty$ , which is an indeterminate form $\frac{\infty}{\infty}$ .

This is still an indeterminate form $\frac{\infty}{\infty}$ .