How to solve the problems of limits?

This page is a practical guide to finding limits of functions. The first thing to try is always substitution. The rest of the page explains what to do when substitution does not give the answer: one-sided limits, the forms $0/0$ and $\infty/\infty,$ the standard limits, L'Hôpital's rule and the other indeterminate forms. Every method comes with solved examples, and there are practice problems with answers at the end.

If limits are new to you, start with Limits (Intuitive Introduction) and Limits and Computational Approach.

On this page: What is a limit · Rules of limits · One-sided limits · The form 0/0 · Standard limits · Limits at infinity · L'Hôpital's rule · Other indeterminate forms · Summary · Practice

What Is a Limit?

The limit of a function at a point $a$ is the number that the values of the function approach as $x$ gets closer and closer to $a.$ In symbols (the equation of the limit):

$$\lim_{x\to a}f(x)=L.$$

It is read "the limit of $f(x)$ as $x$ approaches $a$ is $L$". It means that we can make $f(x)$ as close to $L$ as we like by taking $x$ close enough to $a$ (but ${x\ne a}$).

Only the values near $a$ matter, not the value at $a.$ In the picture the function has some value $f(a)$ at the point $a$ itself, but its values near $a$ approach $L,$ so the limit is $L.$ The function does not even have to be defined at $a.$

A limit can also be infinite: ${\lim_{x\to a}f(x)=+\infty}$ (or $-\infty$) means that the values grow (or decrease) without bound. And $x$ can approach $+\infty$ or $-\infty$ instead of a number. Finally, a limit may not exist at all, as we will see with one-sided limits.

Rules of Limits

If the limits ${\lim_{x\to a}f(x)=L}$ and ${\lim_{x\to a}g(x)=M}$ exist (they are numbers) and $c$ is a constant, then:

Name of the rule	Rule
Constant rule	${\lim_{x\to a}c=c}$
Identity rule	${\lim_{x\to a}x=a}$
Constant multiple rule	${\lim_{x\to a}[c\,f(x)]=c\,L}$
Sum rule	${\lim_{x\to a}[f(x)+g(x)]}={L+M}$
Difference rule	${\lim_{x\to a}[f(x)-g(x)]}={L-M}$
Product rule	${\lim_{x\to a}[f(x)\,g(x)]=L\,M}$
Quotient rule	${\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}},$  if ${M\ne0}$
Power rule	${\lim_{x\to a}[f(x)]^n=L^n},$  ${n=1,2,3,\ldots}$
Root rule	${\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{L}};$ for even $n,$ ${f(x)\ge0}$ near $a$

The same rules hold for the one-sided limits ${x\to a^-},$ ${x\to a^+}$ and, except the identity rule, for ${x\to\pm\infty}.$ They need $L$ and $M$ to be numbers, and the quotient rule needs ${M\ne0}.$ When this fails, use the methods further down the page.

Direct substitution

The rules give the limit of a polynomial at any point $a\text{:}$ it is the value of the polynomial at $a.$ The same is true for a rational function whose denominator is not $0$ at $a.$ The functions $\sin x,$ $\cos x,$ $e^x,$ $\ln x$ and $\sqrt{x}$ behave the same way at every point inside their domains (they are continuous). For all such functions

${\lim_{x\to a}f(x)=f(a)}\text{:}$ just substitute ${x=a}.$

Example 1. Find ${\lim_{x\to1}(4x^2-8x^5+x+12)}.$

First with the rules, step by step:

${\lim_{x\to1}(4x^2-8x^5+x+12)}={\lim_{x\to1}4x^2-\lim_{x\to1}8x^5}+{\lim_{x\to1}x+\lim_{x\to1}12}$ sum and difference rules

$={4\lim_{x\to1}x^2-8\lim_{x\to1}x^5}+{\lim_{x\to1}x+\lim_{x\to1}12}$ constant multiple rule

$={4\cdot1^2-8\cdot1^5+1+12}={9}$ power, identity and constant rules

In practice we skip the middle lines and substitute at once: ${4\cdot1^2-8\cdot1^5+1+12}={9}.$

Example 2. Find the limits.

a) ${\lim_{x\to4}(6x^4+5x^3-4x+16)}={6\cdot4^4+5\cdot4^3}-{4\cdot4+16}={1536+320-16+16}={{1856};}$

b) ${\lim_{x\to2}(12x^5+2\sin x+15x-4)}={12\cdot2^5+2\sin2}+{15\cdot2-4}={410+2\sin2}\approx{411.82}$  (the angle $2$ is in radians);

c) ${\lim_{x\to2}\frac{x^2+1}{x+3}}={\frac{2^2+1}{2+3}}={1},$  because the denominator is not $0$ at ${x=2}.$

You can check answers like these with a limit calculator, but for such functions substitution is the quickest way.

Types of Limits: One-Sided and Two-Sided

There are three basic types of limits at a point $a\text{:}$

• the left-hand limit: $x$ approaches $a$ from the left (${x\lt a}$);
• the right-hand limit: $x$ approaches $a$ from the right (${x\gt a}$);
• the two-sided limit: $x$ approaches $a$ from both sides.

1. Left-hand limit

${\lim_{x\to a^-}f(x)=L}$ means that $f(x)$ approaches $L$ as $x$ approaches $a$ through values smaller than $a.$ The small minus sign in $a^-$ only shows the side; it does not mean that $a$ is negative.

2. Right-hand limit

${\lim_{x\to a^+}f(x)=L}$ means that $f(x)$ approaches $L$ as $x$ approaches $a$ through values larger than $a.$ The small plus sign in $a^+$ only shows the side.

3. Two-sided limit

The ordinary limit ${\lim_{x\to a}f(x)}$ is the two-sided limit. It exists only when both one-sided limits exist and are equal:

${\lim_{x\to a}f(x)=L}$  if and only if  ${\lim_{x\to a^-}f(x)=L}$ and ${\lim_{x\to a^+}f(x)=L}.$

For the functions of the previous section both one-sided limits equal $f(a),$ so all three limits are found by substitution. For example, ${\lim_{x\to4^-}(6x^4+5x^3-4x+16)}={1856},$ just like the two-sided limit in Example 2a. One-sided limits matter where the function behaves differently on the two sides of $a\text{:}$ piecewise functions, absolute values, and fractions whose denominator becomes $0$ at $a.$

Example 3. Find the one-sided limits and the limit of $f$ at ${x=1},$ where

$$f(x)=\begin{cases}x^2+1, & x\lt1,\\ 4-x, & x\ge1.\end{cases}$$

To the left of $1$ the function is ${x^2+1},$ and to the right it is ${4-x}\text{:}$

${\lim_{x\to1^-}f(x)}={\lim_{x\to1^-}(x^2+1)}={{2},}$   ${\lim_{x\to1^+}f(x)}={\lim_{x\to1^+}(4-x)}={{3}.}$

The one-sided limits are different, so ${\lim_{x\to1}f(x)}$ does not exist: the graph jumps at ${x=1}.$ The value ${f(1)=3}$ does not change this. At any other point the limit is found by substitution in the formula that holds near that point, for example ${\lim_{x\to3}f(x)=4-3=1}.$

Absolute values

Example 4. Find ${\lim_{x\to0^+}\frac{\lvert\sin x\rvert}{\sin x}}$ and ${\lim_{x\to0^-}\frac{\lvert\sin x\rvert}{\sin x}}.$

For ${0\lt x\lt\pi}$ we have ${\sin x\gt0},$ so ${\lvert\sin x\rvert=\sin x}$ and the fraction equals $1.$ For ${-\pi\lt x\lt0}$ we have ${\sin x\lt0},$ so ${\lvert\sin x\rvert=-\sin x}$ and the fraction equals $-1.$ Hence

${\lim_{x\to0^+}\frac{\lvert\sin x\rvert}{\sin x}=1},$   ${\lim_{x\to0^-}\frac{\lvert\sin x\rvert}{\sin x}=-1},$

and the two-sided limit at $0$ does not exist.

With an absolute value, find where the expression inside it changes sign, and remove the bars separately on each side.

A nonzero number over zero

If substitution gives ${\frac{c}{0}}$ with ${c\ne0},$ the absolute value of the fraction grows without bound, and the one-sided limits are $+\infty$ or $-\infty.$ The sign of the fraction on each side of $a$ decides which (for a polynomial denominator the sign is the same all along each side, close to $a$).

Example 5. Find ${\lim_{x\to3^-}\frac{x+2}{x-3}},$ ${\lim_{x\to3^+}\frac{x+2}{x-3}}$ and ${\lim_{x\to3}\frac{x+2}{(x-3)^2}}.$

The numerator approaches $5$ and the denominators approach $0.$ For $x$ a little smaller than $3,$ ${x-3}$ is a small negative number, and for $x$ a little larger than $3$ it is a small positive number:

${\lim_{x\to3^-}\frac{x+2}{x-3}=-\infty},$   ${\lim_{x\to3^+}\frac{x+2}{x-3}=+\infty}.$

The square ${(x-3)^2}$ is positive on both sides, so ${\lim_{x\to3}\frac{x+2}{(x-3)^2}=+\infty}.$

The Form $0/0\text{:}$ Simplify First

If substitution gives ${\frac{0}{0}},$ the expression has no value at $a,$ but the limit may still exist. ${\frac{0}{0}}$ is called an indeterminate form: it does not tell you the limit; it only tells you that more work is needed. Rewrite the expression so that substitution works. You may cancel factors that are $0$ at $a,$ because the limit uses only the values ${x\ne a}.$

Factor and cancel

Example 6. Find ${\lim_{x\to2}\frac{4x^2-8x}{4x-8}}.$

Substitution gives ${\frac{16-16}{8-8}=\frac{0}{0}}.$ Both polynomials have the factor ${x-2}\text{:}$

${\lim_{x\to2}\frac{4x^2-8x}{4x-8}}={\lim_{x\to2}\frac{4x(x-2)}{4(x-2)}}={\lim_{x\to2}x}={{2}.}$

Example 7. Find ${\lim_{x\to-1}\frac{x^3+1}{x^2-1}}.$

${\lim_{x\to-1}\frac{x^3+1}{x^2-1}}={\lim_{x\to-1}\frac{(x+1)(x^2-x+1)}{(x+1)(x-1)}}={\lim_{x\to-1}\frac{x^2-x+1}{x-1}}={\frac{3}{-2}}={{-\frac{3}{2}}.}$

If two polynomials are both $0$ at ${x=a},$ both are divisible by ${x-a},$ so this factor can always be cancelled (Horner's rule gives the quotients quickly). If the result is still ${\frac{0}{0}},$ cancel again.

Multiply by the conjugate

Example 8. Find ${\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}}.$

Again ${\frac{0}{0}}.$ Multiply the numerator and the denominator by the conjugate ${\sqrt{x+4}+2}.$ Since ${(\sqrt{x+4}-2)(\sqrt{x+4}+2)}={x+4-4}={{x},}$

${\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}}={\lim_{x\to0}\frac{x}{x(\sqrt{x+4}+2)}}={\lim_{x\to0}\frac{1}{\sqrt{x+4}+2}}={{\frac{1}{4}}.}$

Standard limits

Some limits cannot be found with algebra alone: the first five below have the form ${\frac{0}{0}},$ the last one the form $1^\infty.$ They are proved once and then used as known results (angles in radians):

${\lim_{x\to0}\frac{\sin x}{x}=1}$

${\lim_{x\to0}\frac{\tan x}{x}=1}$

${\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}}$

${\lim_{x\to0}\frac{e^x-1}{x}=1}$

${\lim_{x\to0}\frac{\ln(1+x)}{x}=1}$

${\lim_{x\to0}(1+x)^{1/x}=e}$

The function ${\frac{\sin x}{x}}$ is not defined at $0,$ so its graph has a hole there, at $(0, 1)\text{:}$ its limit at $0$ is $1.$ In every formula of the table, $x$ can be replaced, in every place where it appears, by any expression that approaches $0$ (and is not $0$ near the point). For example, ${\lim_{x\to0}\frac{\sin5x}{5x}=1},$ because ${u=5x}$ approaches $0$ when $x$ does.

Example 9. Find ${\lim_{x\to0}\frac{\sin5x}{2x}}.$

${\lim_{x\to0}\frac{\sin5x}{2x}}={\lim_{x\to0}\left(\frac{5}{2}\cdot\frac{\sin5x}{5x}\right)}={\frac{5}{2}\cdot1}={{\frac{5}{2}}.}$

Example 10. Find ${\lim_{x\to0}\frac{e^{3x}-1}{\sin x}}.$

${\lim_{x\to0}\frac{e^{3x}-1}{\sin x}}={\lim_{x\to0}\left(3\cdot\frac{e^{3x}-1}{3x}\cdot\frac{x}{\sin x}\right)}={3\cdot1\cdot1}={{3}.}$

Limits at Infinity

As ${x\to+\infty}$ or ${x\to-\infty},$ a polynomial behaves like its term of the highest degree. For a quotient of polynomials, substitution gives ${\frac{\infty}{\infty}},$ another indeterminate form. Divide the numerator and the denominator by the highest power of $x$ in the denominator.

Example 11. Find ${\lim_{x\to+\infty}\frac{5x^3-2x}{1-4x^3}}.$

${\lim_{x\to+\infty}\frac{5x^3-2x}{1-4x^3}}={\lim_{x\to+\infty}\frac{5-2/x^2}{1/x^3-4}}={\frac{5-0}{0-4}}={{-\frac{5}{4}}.}$

Example 12. Find ${\lim_{x\to+\infty}\left(\sqrt{x^2+3x}-x\right)}.$

Here substitution gives ${\infty-\infty}.$ Multiply and divide by the conjugate ${\sqrt{x^2+3x}+x},$ then divide by $x$ (for ${x\gt0},$ ${\sqrt{x^2+3x}=x\sqrt{1+3/x}}$):

${\sqrt{x^2+3x}-x}={\frac{(x^2+3x)-x^2}{\sqrt{x^2+3x}+x}}={\frac{3x}{\sqrt{x^2+3x}+x}}={{\frac{3}{\sqrt{1+3/x}+1}}.}$

As ${x\to+\infty},$ this approaches ${\frac{3}{1+1}=\frac{3}{2}}.$ So the limit is ${\frac{3}{2}},$ although "${\infty-\infty}$" might suggest $0.$ More examples are in Limits at Infinity.

How to Solve Limits with L'Hôpital's Rule

L'Hôpital's rule (also written L'Hospital's rule) replaces a limit of the form ${\frac{0}{0}}$ or ${\frac{\infty}{\infty}}$ with the limit of the quotient of the derivatives. To use it you need to know how to find derivatives.

Differentiate the numerator and the denominator separately. This is not the quotient rule for derivatives: the fraction ${\frac{f(x)}{g(x)}}$ itself is not differentiated.

Why it works in the simplest case: if ${f(a)=g(a)=0},$ $f'$ and $g'$ are continuous at $a$ and ${g'(a)\ne0},$ then for ${x\ne a}$

${\frac{f(x)}{g(x)}}={\frac{f(x)-f(a)}{g(x)-g(a)}}={{\frac{\dfrac{f(x)-f(a)}{x-a}}{\dfrac{g(x)-g(a)}{x-a}}},}$

and the two difference quotients approach $f'(a)$ and $g'(a).$ So the limit is ${\frac{f'(a)}{g'(a)}},$ which is ${\lim_{x\to a}\frac{f'(x)}{g'(x)}}$ because $f'$ and $g'$ are continuous.

Example 13. Find ${\lim_{x\to2}\frac{4x^2-8x}{4x-8}}$ with L'Hôpital's rule.

At ${x=2}$ both ${4x^2-8x}$ and ${4x-8}$ are $0,$ so we have the form ${\frac{0}{0}}.$ The derivatives are ${(4x^2-8x)'=8x-8}$ and ${(4x-8)'=4},$ so

${\lim_{x\to2}\frac{4x^2-8x}{4x-8}}={\lim_{x\to2}\frac{8x-8}{4}}={\frac{16-8}{4}}={{2},}$

the same answer as with factoring in Example 6.

Example 14 (the rule used twice). Find ${\lim_{x\to0}\frac{1-\cos x}{x^2}}.$

${\lim_{x\to0}\frac{1-\cos x}{x^2}}={\lim_{x\to0}\frac{\sin x}{2x}}={\lim_{x\to0}\frac{\cos x}{2}}={{\frac{1}{2}}.}$

After the first step the limit is still of the form ${\frac{0}{0}},$ so the rule is used again. (The second step could also use the standard limit ${\lim_{x\to0}\frac{\sin x}{x}=1}.$)

Example 15 (the form ${\frac{\infty}{\infty}}$). Find the limits.

a) ${\lim_{x\to+\infty}\frac{\ln x}{x}}={\lim_{x\to+\infty}\frac{1/x}{1}}={\lim_{x\to+\infty}\frac{1}{x}}={{0};}$

b) ${\lim_{x\to+\infty}\frac{x^2}{e^x}}={\lim_{x\to+\infty}\frac{2x}{e^x}}={\lim_{x\to+\infty}\frac{2}{e^x}}={{0}.}$

Used $n$ times, the rule gives ${\lim_{x\to+\infty}\frac{x^n}{e^x}=0}$ for every $n.$ So $e^x$ grows faster than any power of $x,$ and (by part a) $\ln x$ grows more slowly than $x.$

Other Indeterminate Forms

Besides ${\frac{0}{0}}$ and ${\frac{\infty}{\infty}},$ the indeterminate forms are ${0\cdot\infty},$ ${\infty-\infty},$ $1^\infty,$ $0^0$ and $\infty^0.$ Rewrite them as a fraction, then simplify or use L'Hôpital's rule:

Form	How to rewrite it
${0\cdot\infty}$	${f\cdot g=\frac{f}{1/g}}$  or  ${\frac{g}{1/f}}$
${\infty-\infty}$	a common denominator, or the conjugate
$1^\infty,$ $0^0,$ $\infty^0$	${f^g=e^{g\ln f}}$ (for ${f\gt0}$): find the limit $K$ of ${g\ln f};$ the answer is $e^K$

Example 16 (${0\cdot\infty}$). Find ${\lim_{x\to0^+}x\ln x}.$

As ${x\to0^+},$ ${\ln x\to-\infty}.$ Write the product as a fraction of the form ${\frac{\infty}{\infty}}$ (here ${\frac{-\infty}{+\infty}}$):

${\lim_{x\to0^+}x\ln x}={\lim_{x\to0^+}\frac{\ln x}{1/x}}={\lim_{x\to0^+}\frac{1/x}{-1/x^2}}={\lim_{x\to0^+}(-x)}={{0}.}$

Example 17 (${\infty-\infty}$). Find ${\lim_{x\to0}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)}.$

Bring the fractions to a common denominator. The result has the form ${\frac{0}{0}},$ and L'Hôpital's rule is used twice:

${\lim_{x\to0}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)}={\lim_{x\to0}\frac{e^x-1-x}{x(e^x-1)}}={\lim_{x\to0}\frac{e^x-1}{e^x-1+xe^x}}={\lim_{x\to0}\frac{e^x}{2e^x+xe^x}}={{\frac{1}{2}}.}$

Example 18 ($1^\infty$). Find ${\lim_{x\to+\infty}\left(1+\frac{3}{x}\right)^x}.$

The base approaches $1$ and the exponent grows without bound. Write ${\left(1+\frac{3}{x}\right)^x=e^{x\ln(1+3/x)}}$ and find the limit of the exponent. It has the form ${0\cdot\infty}.$ Written as the fraction ${\frac{\ln(1+3/x)}{1/x}},$ it has the form ${\frac{0}{0}}.$

The derivative of the numerator ${\ln(1+3/x)}$ is ${\frac{-3/x^2}{1+3/x}=-\frac{3}{x^2+3x}},$ and the derivative of the denominator ${1/x}$ is ${-1/x^2}.$ So

${\lim_{x\to+\infty}x\ln\left(1+\frac{3}{x}\right)}={\lim_{x\to+\infty}\frac{\ln(1+3/x)}{1/x}}={\lim_{x\to+\infty}\frac{-3/(x^2+3x)}{-1/x^2}}={\lim_{x\to+\infty}\frac{3x^2}{x^2+3x}}={\lim_{x\to+\infty}\frac{3x}{x+3}}={{3}.}$

The exponent approaches $3,$ so ${\lim_{x\to+\infty}\left(1+\frac{3}{x}\right)^x=e^3}.$

Example 19 ($0^0$). Find ${\lim_{x\to0^+}x^x}.$

${x^x=e^{x\ln x}},$ and ${x\ln x\to0}$ by Example 16. So the limit is ${e^0=1}.$

Summary: How to Solve a Limit Problem

Substitute ${x=a}$ (for ${x\to\pm\infty},$ look at the leading terms) and act on what you get:

You get	What to do
a number, and one formula holds on both sides of $a$	That is the limit (polynomials, rational functions, roots, $\sin x,$ $\cos x,$ $e^x,$ $\ln x$ inside their domains).
a piecewise function or an absolute value: different formulas on the two sides of $a$	Find the left-hand and right-hand limits; the limit exists only if they are equal.
${\frac{c}{0}},$ ${c\ne0}$	The one-sided limits are infinite: $+\infty$ or $-\infty,$ depending on the sign on each side.
${\frac{0}{0}}$	Factor and cancel, multiply by the conjugate, use a standard limit, or L'Hôpital's rule.
${\frac{\infty}{\infty}}$	Divide by the highest power of $x,$ or use L'Hôpital's rule.
${0\cdot\infty},$ ${\infty-\infty}$	Rewrite as one fraction.
$1^\infty,$ $0^0,$ $\infty^0$	Use ${f^g=e^{g\ln f}}.$

Practice Problems

1. Find ${\lim_{x\to2}(3x^3-x+5)}.$

Answer

Substitution: ${3\cdot2^3-2+5}={24-2+5}={{27}.}$

2. Let ${f(x)=2x+1}$ for ${x\lt1}$ and ${f(x)=x^2+3}$ for ${x\ge1}.$ Find ${\lim_{x\to1^-}f(x)}$ and ${\lim_{x\to1^+}f(x)}.$ Does ${\lim_{x\to1}f(x)}$ exist?

Answer

${\lim_{x\to1^-}(2x+1)=3}$ and ${\lim_{x\to1^+}(x^2+3)=4}.$ They are different, so the limit at $1$ does not exist.

3. Find ${\lim_{x\to-2}\frac{x^2-4}{x^2+5x+6}}.$

Answer

Substitution gives ${\frac{0}{0}}.$ Factor: ${\frac{(x-2)(x+2)}{(x+2)(x+3)}=\frac{x-2}{x+3}},$ so the limit is ${\frac{-4}{1}=-4}.$

4. Find ${\lim_{x\to9}\frac{\sqrt{x}-3}{x-9}}.$

Answer

${x-9=(\sqrt{x}-3)(\sqrt{x}+3)},$ so the fraction is ${\frac{1}{\sqrt{x}+3}}$ and the limit is ${\frac{1}{6}}.$

5. Find ${\lim_{x\to0}\frac{\sin3x}{\sin7x}}.$

Answer

${\frac{\sin3x}{\sin7x}}={\frac{3}{7}\cdot\frac{\sin3x}{3x}\cdot\frac{7x}{\sin7x}},$ so the limit is ${\frac{3}{7}}.$

6. Find ${\lim_{x\to0}\frac{x-\sin x}{x^3}}.$

Answer

L'Hôpital's rule three times: ${\lim_{x\to0}\frac{x-\sin x}{x^3}}={\lim_{x\to0}\frac{1-\cos x}{3x^2}}={\lim_{x\to0}\frac{\sin x}{6x}}={\lim_{x\to0}\frac{\cos x}{6}}={{\frac{1}{6}}.}$

7. Find ${\lim_{x\to+\infty}\frac{(\ln x)^2}{x}}.$

Answer

The form ${\frac{\infty}{\infty}};$ L'Hôpital's rule twice: ${\lim_{x\to+\infty}\frac{(\ln x)^2}{x}}={\lim_{x\to+\infty}\frac{2\ln x}{x}}={\lim_{x\to+\infty}\frac{2}{x}}={{0}.}$

8. Find ${\lim_{x\to0^+}x^2\ln x}.$

Answer

${x^2\ln x=\frac{\ln x}{1/x^2}}$ (the form ${\frac{\infty}{\infty}}$), so ${\lim_{x\to0^+}\frac{\ln x}{1/x^2}}={\lim_{x\to0^+}\frac{1/x}{-2/x^3}}={\lim_{x\to0^+}\left(-\frac{x^2}{2}\right)}={{0}.}$

9. Find ${\lim_{x\to+\infty}\left(\sqrt{x^2+x}-x\right)}.$

Answer

With the conjugate: ${\sqrt{x^2+x}-x}={\frac{x}{\sqrt{x^2+x}+x}}={{\frac{1}{\sqrt{1+1/x}+1}},}$ which approaches ${\frac{1}{2}}.$

10. Find ${\lim_{x\to0}(1+2x)^{1/x}}.$

Answer

${(1+2x)^{1/x}=e^{\ln(1+2x)/x}}$ and ${\lim_{x\to0}\frac{\ln(1+2x)}{x}}={\lim_{x\to0}\frac{2}{1+2x}}={2}$ (L'Hôpital's rule), so the limit is $e^2.$

See also: Limits (Intuitive Introduction), Limits and Computational Approach, Limits of sequences, Properties of continuous functions and the derivative calculator.