LIMITS

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$$f(x)=\frac{x^2-1}{x-1}$$

Let evaluate f(1)

$$f(1)=\frac{(1)^2-1}{1-1}$$

$f(1)=\frac{0}{0}$ which is undefined.

In a situation like this, we can only determine the value of $f(1) as x tends to approach 1. This is the idea of limits

The limit of a function shows the value of a function when its input approaches a particular value.

There are three major ways of solving for limits although other methods exist.

There are:

1. Direct substitution: Simple inserting the value 

2. Factorization method: By factorizing

3. L'hopital method: Taking the derivative of the numerator and dividing by the derivative of the denominator 

Returning to our earlier example, 

$f(x)=\frac{x^2-1}{x-1}$, we can evaluate the limits as x tends to approach 1 instead of $f(1)$. That is,

$$\lim_{x \to 1} \frac{x^2-1}{x-1}$$

As we saw earlier, using direct substitution will lead to an undefined fraction and we don't want that, so, we use the factorization method.

$$\lim_{x \to 1} \frac{x^2-1}{x-1}$$

$x^2-1$ is a difference of two squares

$$\lim_{x \to 1}\frac{(x+1)\require{cancel}\bcancel{(x-1)}}{\require{cancel}\bcancel{(x-1)}}$$

$$\lim_{x \to 1}x+1$$

Now, we can use direct substitution

$$\lim_{x \to 1}1+1=2$$

Example 1

Evaluate $\lim_{x \to 3}10x-4$

Solution:

$$\lim_{x \to 3}10x-4$$

By direct substitution 

$\lim_{x \to 3}10(3)-4=26$

Example 2

Evaluate $\lim_{x \to 10}\frac{x^2-100}{x-10}$

Solution

If we use direct substitution, we will get an undefined fraction because the denominator will be zero.

Let's factorize

$$\lim_{x \to 10}\frac{(x+10)(x-10)}{x-10}$$

$$\lim_{x \to 10}x+10$$

Now, we can use direct substitution

$\lim_{x \to 10}10+10=20$

Example 3

Evaluate $\lim_{x \to 8}\frac{\sqrt[3]{x}-2}{x-8}$

Solution:

If we use direct substitution, we will get an undefined fraction. This expression is not factorizable. Hence, we will use the l'hopital rule.

L'hopital rule says we should take the derivative of the numerator and take the derivative of the denominator.

The derivative of the numerator $\sqrt[3]{x}-2$ is $\frac{1}{3(\sqrt[3]{x})^2}$

The derivative of the denominator ($x-8$) is 1.

Therefore

$$\lim_{x \to 8}\frac{\frac{1}{3(\sqrt[3]{x})^2}}{1}$$

$$\lim_{x \to 8}\frac{1}{3(\sqrt[3]{x})^2}$$

Using direct substitution

$$\lim_{x \to 8}\frac{1}{3(\sqrt[3]{8})^2}$$

$$\lim_{x \to 8}\frac{1}{3(4)}$$

$$\lim_{x \to 8}\frac{1}{12}$$

Example 4

Evaluate $\lim_{x \to 16}\frac{\sqrt{x}-4}{x-16}$

Solution:

$\lim_{x \to 16}\frac{(x)^½-4}{x-16}$

Direct substitution and factorization won't work, so we use the hospital rule.

Like we did earlier, we take the derivative of the numerator and denominator.

The derivative of the numerator is $\frac{1}{2\sqrt{x}}$

The derivative of the denominator is 1

$$\lim_{x \to 16}\frac{\frac{1}{2\sqrt{x}}}{1}$$

$$\lim_{x \to 16}\frac{1}{2\sqrt{x}}$$

Now, let apply the limit expression

$$\lim_{x \to 16}\frac{1}{2\sqrt{16}}$$

$$\lim_{x \to 16}\frac{1}{2(4)}=\frac{1}{8}$$

That will be all for now. But before you go, here are other related posts that you might want to try out.

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