Evaluate the following limits: x infinity ( x)/(sqrt(x)) and x 0 ( x)/(x)

Question

Asked on May 4, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-05-04T16:59:00.576Z

You're on a roll — let's evaluate these limits. 1) _x ( x)/(sqrt(x)) Step 1: Check for indeterminate form. As x , x and sqrt(x) . This is an indeterminate form of type ()/(), so we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule. Take the derivative of the numerator and the denominator: (d)/(dx)( x) = (1)/(x) (d)/(dx)(sqrt(x)) = (d)/(dx)(x^1/2) = (1)/(2)x^-1/2 = (1)/(2sqrt(x)) Now, evaluate the limit of the ratio of these derivatives: _x (1)/(x)(1)/(2sqrt(x)) = _x (1)/(x) · 2sqrt(x) Step 3: Simplify and evaluate the limit. _x 2sqrt(x)x = _x (2)/(x^1/2) = _x (2)/(sqrt(x)) As x , sqrt(x) , so (2)/(sqrt(x)) 0. The limit is 0. 2) _x 0 ( x)/(x) Step 1: Check for indeterminate form. As x 0, x 0 = 0 and x 0. This is an indeterminate form of type (0)/(0), so we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule. Take the derivative of the numerator and the denominator: (d)/(dx)( x) = ^2 x (d)/(dx)(x) = 1 Now, evaluate the limit of the ratio of these derivatives: _x 0 (^2 x)/(1) Step 3: Evaluate the limit. Substitute x=0: (^2 0)/(1) = (1)/(( 0)^2) = (1)/(1^2) = 1 The limit is 1. 3) _x (x^2)/(e^x) Step 1: Check for indeterminate form. As x , x^2 and e^x . This is an indeterminate form of type ()/(), so we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule (first time). Take the derivative of the numerator and the denominator: (d)/(dx)(x^2) = 2x (d)/(dx)(e^x) = e^x The new limit is _x (2x)/(e^x). This is still an indeterminate form of type ()/(). Step 3: Apply L'Hôpital's Rule (second time). Take the derivative of the new numerator and denominator: (d)/(dx)(2x) = 2 (d)/(dx)(e^x) = e^x The new limit is _x (2)/(e^x). Step 4: Evaluate the limit. As x , e^x , so (2)/(e^x) 0. The limit is 0. 4) _x 0 ((1+x))/(x) Step 1: Check for indeterminate form. As x 0, (1+x) (1) = 0 and x 0. This is an indeterminate form of type (0)/(0), so we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule. Take the derivative of the numerator and the denominator: (d)/(dx)((1+x)) = (1)/(1+x) (d)/(dx)(x) = 1 Now, evaluate the limit of the ratio of these derivatives: _x 0 (1)/(1+x)1 Step 3: Evaluate the limit. Substitute x=0: (1)/(1+0) = (1)/(1) = 1 The limit is 1. Got more? Send 'em.

Accepted Answer · 2026-05-04T16:59:00.576Z

You're on a roll — let's evaluate these limits. 1) _x ( x)/(sqrt(x)) Step 1: Check for indeterminate form. As x , x and sqrt(x) . This is an indeterminate form of type ()/(), so we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule. Take the derivative of the numerator and the denominator: (d)/(dx)( x) = (1)/(x) (d)/(dx)(sqrt(x)) = (d)/(dx)(x^1/2) = (1)/(2)x^-1/2 = (1)/(2sqrt(x)) Now, evaluate the limit of the ratio of these derivatives: _x (1)/(x)(1)/(2sqrt(x)) = _x (1)/(x) · 2sqrt(x) Step 3: Simplify and evaluate the limit. _x 2sqrt(x)x = _x (2)/(x^1/2) = _x (2)/(sqrt(x)) As x , sqrt(x) , so (2)/(sqrt(x)) 0. The limit is 0. 2) _x 0 ( x)/(x) Step 1: Check for indeterminate form. As x 0, x 0 = 0 and x 0. This is an indeterminate form of type (0)/(0), so we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule. Take the derivative of the numerator and the denominator: (d)/(dx)( x) = ^2 x (d)/(dx)(x) = 1 Now, evaluate the limit of the ratio of these derivatives: _x 0 (^2 x)/(1) Step 3: Evaluate the limit. Substitute x=0: (^2 0)/(1) = (1)/(( 0)^2) = (1)/(1^2) = 1 The limit is 1. 3) _x (x^2)/(e^x) Step 1: Check for indeterminate form. As x , x^2 and e^x . This is an indeterminate form of type ()/(), so we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule (first time). Take the derivative of the numerator and the denominator: (d)/(dx)(x^2) = 2x (d)/(dx)(e^x) = e^x The new limit is _x (2x)/(e^x). This is still an indeterminate form of type ()/(). Step 3: Apply L'Hôpital's Rule (second time). Take the derivative of the new numerator and denominator: (d)/(dx)(2x) = 2 (d)/(dx)(e^x) = e^x The new limit is _x (2)/(e^x). Step 4: Evaluate the limit. As x , e^x , so (2)/(e^x) 0. The limit is 0. 4) _x 0 ((1+x))/(x) Step 1: Check for indeterminate form. As x 0, (1+x) (1) = 0 and x 0. This is an indeterminate form of type (0)/(0), so we can apply L'Hôpital's Rule. Step 2: Apply L'Hôpital's Rule. Take the derivative of the numerator and the denominator: (d)/(dx)((1+x)) = (1)/(1+x) (d)/(dx)(x) = 1 Now, evaluate the limit of the ratio of these derivatives: _x 0 (1)/(1+x)1 Step 3: Evaluate the limit. Substitute x=0: (1)/(1+0) = (1)/(1) = 1 The limit is 1. Got more? Send 'em.

Evaluate the following limits: x infinity ( x)/(sqrt(x)) and x 0 ( x)/(x)

1) $\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$

2) $\lim_{x \to 0} \frac{\tan x}{x}$

3) $\lim_{x \to \infty} \frac{x^2}{e^x}$

4) $\lim_{x \to 0} \frac{\ln(1+x)}{x}$

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Evaluate the following limits: x infinity ( x)/(sqrt(x)) and x 0 ( x)/(x)

1) $\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$

2) $\lim_{x \to 0} \frac{\tan x}{x}$

3) $\lim_{x \to \infty} \frac{x^2}{e^x}$

4) $\lim_{x \to 0} \frac{\ln(1+x)}{x}$

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Evaluate the following limits: x infinity ( x)/(sqrt(x)) and x 0 ( x)/(x)

1) lim⁡x→∞ln⁡xx\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}limx→∞​x​lnx​

2) lim⁡x→0tan⁡xx\lim_{x \to 0} \frac{\tan x}{x}limx→0​xtanx​

3) lim⁡x→∞x2ex\lim_{x \to \infty} \frac{x^2}{e^x}limx→∞​exx2​

4) lim⁡x→0ln⁡(1+x)x\lim_{x \to 0} \frac{\ln(1+x)}{x}limx→0​xln(1+x)​

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Evaluate the following limits: x infinity ( x)/(sqrt(x)) and x 0 ( x)/(x)

1) lim⁡x→∞ln⁡xx\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}limx→∞​x​lnx​

2) lim⁡x→0tan⁡xx\lim_{x \to 0} \frac{\tan x}{x}limx→0​xtanx​

3) lim⁡x→∞x2ex\lim_{x \to \infty} \frac{x^2}{e^x}limx→∞​exx2​

4) lim⁡x→0ln⁡(1+x)x\lim_{x \to 0} \frac{\ln(1+x)}{x}limx→0​xln(1+x)​

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1) $\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$

2) $\lim_{x \to 0} \frac{\tan x}{x}$

3) $\lim_{x \to \infty} \frac{x^2}{e^x}$

4) $\lim_{x \to 0} \frac{\ln(1+x)}{x}$

1) $\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$

2) $\lim_{x \to 0} \frac{\tan x}{x}$

3) $\lim_{x \to \infty} \frac{x^2}{e^x}$

4) $\lim_{x \to 0} \frac{\ln(1+x)}{x}$