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## Calculus - Basic Limits problems

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### Calculus - Basic Limits problems

Finding limits of functions or expressions is an important part of mathematics calculus. let's learn by examples.

Find:

1. $\lim_{x\to 5} \frac{x^2 - 25}{x-5}$

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

3. $\lim_{x\to \infty}\frac{3x^2 - 6}{4x - 8}$

4. $\lim \frac{5x^3 + 8x^2}{3x^3 - 16x^2}$

5. $\lim_{x\to 2} \frac{x^3 - 8}{x - 2}$

6. $\lim_{x \to 1} \frac{x^3-1}{x^2 - 1}$

Solution 1.

$\lim_{x\to 5} \frac{x^2 - 25}{x-5} = \lim_{x\to 5} \frac{(x+5)(x-5)}{x-5} = \lim_{x\to 5} (x + 5) = 10$

Solution 2.

$\lim_{x\to 1} \frac{x^2 -x -2}{x-1} = \lim_{x\to 1} \frac{(x+1)(x-2)}{(x+1)(x-1)} = \lim_{x\to 1} \frac{x-2}{x-1} = -\frac{1}{0}$, undefined.

Solution 3.

$\lim_{x\to \infty} \frac{3x^2 - 6x}{4x -8} = \lim_{x\to \infty} \frac{3x^2 - 6x}{4x -8}\left( \dfrac{\frac{1}{x^2}}{\frac{1}{x^2}}\right) = \lim_{x\to \infty} \frac{3- \frac{6}{x}}{\frac{4}{x}-\frac{8}{x^2}} = \frac{3-0}{0-0} = \frac{3}{0}$, the limit is undefined.

Again,

$\lim_{x\to \infty} \frac{3x^2 - 6x}{4x -8} = \lim_{x\to \infty}\frac{3x(x-2)}{4(x-2)} = \lim_{x\to \infty} \frac{3x}{4} \to \infty$

Hence there is no limit to this problem.

Solution 4.

$\lim_{x\to0} \frac{5x^3 + 8x^2}{3x^3 - 16x^2} = \lim_{x\to 0} \frac{5x^3 + 8x^2}{3x^3 - 16x^2}\left( \dfrac{\frac{1}{x^2}}{\frac{1}{x^2}}\right) = \lim_{x\to 0} \frac{5x + 8}{3x - 16} = -\frac{8}{16} = -\frac{1}{2}$

Solution 5.

$\lim_{x\to 2} \frac{x^3 - 8}{x - 2} = \lim_{x\to 2} \frac{(x-2)(x^2 + 2x + 4)}{x-2} = \lim_{x\to 2} (x^2 + 2x + 4) = 4 + 4 + 4 = 12$

Solution 6.

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

Now, find:

6. (a) $\lim_{x\to 5} \frac{x^2 - 25}{x+5}$

(b) $\lim_{x\to 5} \frac{x + 5}{x^2-25}$

7. Show that $\lim_{x\to 4} \frac{2- \sqrt{x}}{4-x} = \frac{1}{4}$

Post more problems and answers to help others learn.
$\Huge \int$ d