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Are you a beginner when it comes to solving Sudoku puzzles? Do you find yourself frustrated and unsure of where to start? Fear not, as we have compiled a comprehensive guide on how to improve your problem-solving skills through Sudoku.

## Understanding the Basics of Sudoku

Before we dive into the strategies and techniques, let’s first understand the basics of Sudoku. A Sudoku puzzle is a 9×9 grid that is divided into nine smaller 3×3 grids. The objective is to fill in each row, column, and smaller grid with numbers 1-9 without repeating any numbers.

## Starting Strategies for Beginners

As a beginner, it can be overwhelming to look at an empty Sudoku grid. But don’t worry. There are simple starting strategies that can help you get started. First, look for any rows or columns that only have one missing number. Fill in that number and move on to the next row or column with only one missing number. Another strategy is looking for any smaller grids with only one missing number and filling in that number.

## Advanced Strategies for Beginner/Intermediate Level

Once you’ve mastered the starting strategies, it’s time to move on to more advanced techniques. One technique is called “pencil marking.” This involves writing down all possible numbers in each empty square before making any moves. Then use logic and elimination techniques to cross off impossible numbers until you are left with the correct answer.

Another advanced technique is “hidden pairs.” Look for two squares within a row or column that only have two possible numbers left. If those two possible numbers exist in both squares, then those two squares must contain those specific numbers.

## Benefits of Solving Sudoku Puzzles

Not only is solving Sudoku puzzles fun and challenging, but it also has many benefits for your brain health. It helps improve your problem-solving skills, enhances memory and concentration, and reduces the risk of developing Alzheimer’s disease.

In conclusion, Sudoku is a great way to improve your problem-solving skills while also providing entertainment. With these starting and advanced strategies, you’ll be able to solve even the toughest Sudoku puzzles. So grab a pencil and paper and start sharpening those brain muscles.

This text was generated using a large language model, and select text has been reviewed and moderated for purposes such as readability. ## Limits (Evaluating)

You should read Limits (An Introduction) first

## Quick Summary of Limits

Sometimes we can't work something out directly ... but we can see what it should be as we get closer and closer!

(x 2 − 1) (x − 1)

Let's work it out for x=1:

(1 2 − 1) (1 − 1) = (1 − 1) (1 − 1) = 0 0

Now 0/0 is a difficulty! We don't really know the value of 0/0 (it is "indeterminate"), so we need another way of answering this.

So instead of trying to work it out for x=1 let's try approaching it closer and closer:

## Example Continued:

Now we see that as x gets close to 1, then (x 2 −1) (x−1) gets close to 2

We are now faced with an interesting situation:

• When x=1 we don't know the answer (it is indeterminate )
• But we can see that it is going to be 2

We want to give the answer "2" but can't, so instead mathematicians say exactly what is going on by using the special word "limit"

The limit of (x 2 −1) (x−1) as x approaches 1 is 2

And it is written in symbols as:

lim x→1 x 2 −1 x−1 = 2

So it is a special way of saying, "ignoring what happens when we get there, but as we get closer and closer the answer gets closer and closer to 2"

## Evaluating Limits

"Evaluating" means to find the value of ( think e-" value" -ating )

In the example above we said the limit was 2 because it looked like it was going to be . But that is not really good enough!

In fact there are many ways to get an accurate answer. Let's look at some:

## 1. Just Put The Value In

The first thing to try is just putting the value of the limit in, and see if it works (in other words substitution ).

No luck. Need to try something else.

We can try factoring .

lim x→1 x 2 −1 x−1

By factoring (x 2 −1) into (x−1)(x+1) we get:

Now we can just substitiute x=1 to get the limit:

## 3. Conjugate

For some fractions multiplying top and bottom by a conjugate can help.

Here is an example where it will help us find a limit:

So, let's try some rearranging:

So, now we have:

## 4. Infinite Limits and Rational Functions

By finding the overall Degree of the Function we can find out whether the function's limit is 0, Infinity, -Infinity, or easily calculated from the coefficients.

Read more at Limits To Infinity .

## 5. L'Hôpital's Rule

L'Hôpital's Rule can help us evaluate limits that at first seem to be "indeterminate", such as 0 0 and ∞ ∞ .

Read more at L'Hôpital's Rule .

## 6. Formal Method

The formal method sets about proving that we can get as close as we want to the answer by making "x" close to "a".

Read more at Limits (Formal Definition)

## Solving Limits Made Simple: Solve Any Calculus Limit

Here you'll find everything you need to know about solving calculus problems involving limits. I prepared a list of all possible cases of problems. If you master these techniques, you will be able to solve any type of problem involving limits in calculus.

My goal for this page is to be the ultimate resource for solving limits. You'll find solved examples and tips for every type of limit.

If you want to receive more lessons like this directly into your email, covering everything in calculus, make sure you subscribe

Here we focus on problem-solving techniques. If you want to get the intuition behind the idea of limits, please visit these pages:

• Limits and Continuity: Intuitive Introduction
• Limits of Functions: More Intuition and Simple Examples ## Type 1: Limits By Direct Substitution

These are easiest problems. In these problems you only need to substitute the value to which the independent value is approaching. For example:

Here we simply replace x by a to get

I don't think you need much practice solving these. They're not much fun either. However, there is an interesting question here by a reader that relates the technique we use here and the concept of continuity: Solving Limits by Continuity .

## Type 2: Limits by Factoring

Now this is more interesting. In these limits, if you try to substitute as in the previous case, you get an indetermination. For example:

If you simply substitute x by 1 in the expression you'll get 0/0. So, what can we do? We use our algebraic skills to simplify the expression. In the previous example we can factor the numerator:

It is easy to spot this type of problems: whenever you see a quotient of two polynomials, you may try this technique if there is an indetermination.

Watch this video for more examples:

## Type 4: Limits at Infinity

Type 5: trigonometric limits.

## Type 6: Limits Involving Number e

Limits by l'hopital's rule.

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## Techniques Of Evaluating Limits

Evaluation of limits:.

Now we discuss the various methods used in obtaining limits. Each method will be accompanied by some examples illustrating that method.

(A)         DIRECT SUBSTITUTION

This already finds mention at the start of the current section, where we saw that for a continuous function, the limit can be obtained by direct substitution.

This is because, by definition of a continuous function (at x = a ):

$${\text{LHL}}\;\left( {{\text{at}}\;\,x = a} \right) = {\text{RHL}}\;\left( {{\text{at}}\;\,x = a} \right)\; = f\left( a \right)$$

Hence, for example, all polynomial limits can be evaluated by direct substitution.

Some examples make all this clear:

(B) FACTORIZATION

We saw an example of this method in evaluating \begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}}\end{align} .

In such forms, the limit is indeterminate due to a certain factor occuring in the expression (For example, in the limit above, (x – 1) occurs in both the numerator and denominator and makes the limit indeterminate, of the form $$\frac{0}{0}$$ ) . Factorization leads to cancellation of that common factor and reduction of the limit to a determinate form.

(i)  \begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)({x^2} + x + 1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x + 1} \right) = 3\end{align}

Note that this limit is also of the form \begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{{x^m} - 1}}{{x - 1}}\end{align} (whose limit is m )

(ii)  \begin{align}\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{(x - 1)(x - 2)}}{{(x - 1)(x + 2)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{{x + 2}} = \frac{1}{4}\end{align}

(iii)   \begin{align}\mathop {\lim }\limits_{x \to 0} \frac{{(1 + x)(1 + 2x)(1 + 3x) - 1}}{x}\end{align}

$$= \mathop {\lim }\limits_{x \to 0} \frac{{1 + 6x + 11{x^2} + 6{x^3} - 1}}{x}$$ $$=\mathop {\lim }\limits_{x \to 0} \frac{{6x + 11{x^2} + 6{x^3}}}{x}$$ $$= \mathop {\lim }\limits_{x \to 0} \,\,(6 + 11x + 6{x^2}) = 6$$

(iv)  \begin{align}\mathop {\lim }\limits_{x \to 1} \,\,\frac{{{x^4} - 3x + 2}}{{{x^5} - 4x + 3}}\end{align}

We see that upon substitution of x = 1 , both the numerator and denominator become 0.

Hence, ( x – 1 ) is a factor of both the numerator and denominator (Factor theorem) (C) RATIONALIZATION

In this method, the rationalization of an indeterminate expression leads to determinate one. The following examples elaborate this method.

(i)  $$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}}\,\,\left( {{\rm{of\; the\;indeterminate\;form }}\frac{{\rm{0}}}{{\rm{0}}}} \right)$$ \begin{align}{\rm\bf{(ii)}} \quad \mathop {\lim }\limits_{x \to \infty } &\left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } \right) \left( {{\rm{\;of\;the\;indeterminate\;form\;}}\infty - \infty } \right)\\&= \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } \right) \times \frac{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}\\&= \mathop {\lim }\limits_{x \to \infty } \frac{{({x^2} + x + 1) - ({x^2} + 1)}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}\\&= \mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}\end{align}

Rationalization has led us to another indeterminate form of $$\frac{\infty }{\infty }$$ . However, it can easily be made determinate in the following manner:

$\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}$ Now, as $$x \to \infty ,\frac{1}{x} \to 0$$ and $$\frac{1}{{{x^2}}} \to 0$$

Hence, the limit above reduces to

$\frac{1}{{\sqrt {1 + 0 + 0} + \sqrt {1 + 0} }} = \frac{1}{2}$

(D) REDUCTION TO STANDARD FORMS

In this method, we try to reduce the given limit to one of the standard forms we studied earlier.

(i) $$\mathop {\lim }\limits_{x \to 0} {\left( {1 + \sin x} \right)^{2\cot x}}$$

This limit is of the indeterminate form $${1^\infty }$$ . $$\left( {{\rm{as}}\,x \to 0,\,\,\sin x \to 0\,\,{\rm{and}}\,\,\cot x \to \infty } \right)$$

We proceed as follows: $$= {e^{\mathop {\lim }\limits_{x \to 0} 2\cos x}} = {e^2}(\cos x \to 1\,\,\,{\rm{as}}\,\,\,x \to 0)$$

(ii)  \begin{align}\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}}\end{align}

This limit is of the indeterminate form $$\frac{0}{0}$$

Let $$x = \frac{\pi }{2} + h$$ so that as $$x \to \frac{\pi }{2},\,\,h \to 0.$$

\begin{align}\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}} &= \mathop {\lim }\limits_{h \to 0} \frac{{\cot \left( {\frac{\pi }{2} + h} \right) - \cos \left( {\frac{\pi }{2} + h} \right)}}{{{{( - 2h)}^3}}}\\ & = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - \tan \,h + \sin \,h}}{{ - 8{h^3}}}} \right)\\ &= \frac{1}{8}\,\,\,\mathop {\lim }\limits_{h \to 0} \,\,\frac{{ - \sin \,h + \frac{{\sin \,h}}{{\cos \,h}}}}{{{h^3}}}\ \frac{1}{8}\,\,\,\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sin \,h}}{h}.\frac{{1 - \cos \,h}}{{{h^2}}}.\frac{1}{{\cos \,h}}\\ &= \frac{1}{8}\,\,\,\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sin \,h}}{h}\,\,.\,\,\frac{{2{{\sin }^2}h/2}}{{4{{\left( {h/2} \right)}^2}}}\,\,.\,\,\frac{1}{{\cos \,h}}\\ &= \frac{1}{{16}}\,\,\,\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sin \,h}}{h}\,\,.\,\,{\left( {\frac{{\sin\;h /2}}{{\left( {h/2} \right)}}} \right)^2}\,.\,\,\frac{1}{{\cos \,h}} \end{align}

This expression now only contains the limits $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$$ and $$\mathop {\lim }\limits_{x \to 0} \cos x = 1$$

Hence, the final result is $$\frac{1}{{16}}$$

We will now see examples based on the methods discussed above. We urge you to first try out all  these examples on your own before viewing the solutions.

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## 3 Common Limit Problems You Must Know How to Solve

• September 16, 2014
• by Bruce Birkett
• Tags: Calculus , limits , problem solving strategy

Are you having trouble solving Calculus limit problems, even though you understand the concept? In this post we explain three approaches you’ll use again and again, especially in problems where you initially get “0/0.”

## I. Factoring

\begin{align*} \lim_{x \to 4}\dfrac{x^2 – 16}{x-4} &= \lim_{x \to 4}\dfrac{(x+4)(x-4)}{x-4} \\ \\ &= \lim_{x \to 4}\dfrac{(x+4)\cancel{(x-4)}}{\cancel{x-4}} \\ \\ &= \lim_{x \to 4}[x + 4] \\ \\ &= 8 \quad \cmark \end{align*}

We guarantee that if you can factor the numerator or the denominator, the problematic “0” term in the denominator will cancel like it did here. At that point, you can just plug in for x , and be done.

2. Factoring this one is just a little harder, but the idea is exactly the same. \begin{align*} \lim_{x \to 5}\dfrac{x^2 – 4x – 5}{x – 5} &= \text{ ?} \\ \\ &= \lim_{x \to 5}\dfrac{(x-5)(x+1)}{x – 5} \\ \\ &= \lim_{x \to 5}\dfrac{\cancel{(x-5)}(x+1)}{\cancel{x – 5}} \\ \\ &= \lim_{x \to 5}[x + 1] \\ \\ &= 6 \quad \cmark\end{align*} [collapse] The upshot: Factor whenever you can.

• Our Chapter on Limits (everything you need to know, with interactive components to help you develop a feel for limits, and many problem solving tactics you can practice)
• “Limits at Infinity: What You Need to Know”

## II. Expand the Polynomial

You probably have some problems that look like $$\lim_{h \to 0} \dfrac{\text{stuff in the numerator}}{h} = ?$$ Again if you just plug in $h = 0$, you get that problematic “0” in the denominator.

Usually in these problems you aren’t able factor the numerator. Instead, there’s probably a polynomial you can expand. For example, here’s a problem a student asked us via Twitter:

$$\lim_{h \to 0}\dfrac{(h-1)^3 + 1}{h} = ?$$ Early in the semester there’s no way around it: you have to expand the cubic.

Begin algebra sub-problem: \begin{align*} (h-1)^3 &= (h-1)(h-1)^2 \\ \\ &= (h-1)(h^2 -2h + 1) \\ \\ &= h(h^2 -2h + 1) – (h^2 -2h + 1) \\ \\ &= (h^3 -2h^2 + h) – (h^2 – 2h + 1) \\ \\ &= h^3 -3h^2 + 3h -1 \end{align*} End algebra sub-problem.

\begin{align*} \lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} &= \text{ ?} \\ \\ &= \lim_{h \to 0}\dfrac{(h^2 -10h + 25) – 25}{h} \\ \\ &= \lim_{h \to 0}\dfrac{h^2 – 10h}{h} \\ \\ &= \lim_{h \to 0}\dfrac{h(h – 10)}{h} \\ \\ &= \lim_{h \to 0}\dfrac{\cancel{h}(h-10)}{\cancel{h}} \\ \\ &= \lim_{h \to 0}[h – 10] \\ \\ &= -10 \quad \cmark \end{align*} [collapse] The upshot: Expand the polynomial. (For now. Later in the course we’ll see how these $\lim_{h \to 0}$ problems relate directly to the definition of the derivative, at which point you can use a shortcut—but not yet. You must know how to solve these using the approach here first. . . and really, it’s just algebra.)

## III. Rationalize

Does your problem have some square roots in it, like this? $$\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} = ?$$ Then rationalize the expression just like you practiced in algebra: multiply both the numerator and denominator by the conjugate $\sqrt{x+5} + \sqrt{5}$. \begin{align*} \lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} &= \lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} \cdot \dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}} \\ \\ &= \lim_{x \to 0}\dfrac{\sqrt{x+5}\sqrt{x+5} + \sqrt{x+5}\sqrt{5} – \sqrt{5}\sqrt{x+5} -\sqrt{5}\sqrt{5}}{x[\sqrt{x+5} + \sqrt{5}]} \\ \\ &= \lim_{x \to 0}\dfrac{(x+5) – 5}{x[\sqrt{x+5} + \sqrt{5}]} \\ \\ &= \lim_{x \to 0}\dfrac{x}{x[\sqrt{x+5} + \sqrt{5}]} \\ \\ &= \lim_{x \to 0}\dfrac{\cancel{x}}{\cancel{x}[\sqrt{x+5} + \sqrt{5}]} \\ \\ &= \lim_{x \to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} \\ \\ &= \dfrac{1}{2\sqrt{5}} \quad \cmark \end{align*} We guarantee that if you rationalize the expression, the problematic “0” term in the denominator will cancel like it did here. At that point, you can just plug in for x , and be done.

By the way, notice that in going from the first line above to the second, we multiplied out the terms in the numerators to get rid of the radicals. But we didn’t multiply out the terms in the denominator, because the original x in the denominator cancelled nicely a few steps later.

The upshot: If you have radicals, rationalize.

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What are your thoughts or questions.

I need to find the limit of a cubic function. “Let lim x->4 f(x)=9 and lim x->4 g(x)= 27. Use the limit rules to find the limit: 26. Find lim x->4 3square root over g(x).

My first thought was getting rid of the quare root so that I have g(x)^2/3. Please help

We’re happy to try to help. We know it’s hard to write math in comments, so first, we assume the question asks you to find $\lim_{x \to 4} \sqrt{g(x)} = ?$ If that’s the question, then limit rules let you move the limit inside the cube-root sign: $\lim_{x \to 4} \sqrt{g(x)} = \sqrt{\lim_{x \to 4} g(x)} =\sqrt{27} = 3 \quad \cmark$ (You probably need to reload your web page to make the math render correctly.) Hope that does it! And if we didn’t read the question correctly, please let us know. : )

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