# Math Help - limits

1. ## limits

Hello
I want to learn all methods for solving limits
can someone help me?

3. I have been searching it in google for about 3 days but I can't find anything good about limits that includes all method, if you have link please share it

4. Well here's a simple one, the method is called direct substitution.

If possible, just substitute the value into the function

$\displaystyle \lim_{x\to 0 } 3x+5 = 3(0)+5= 5$

Now you have a go

$\displaystyle \lim_{x\to 2 } 4x^2+2x-1 = \dots$

$\displaystyle \lim_{x\to 1 } \frac{x+3}{x^2+x-8} = \dots$

Then try this one, what happens?

$\displaystyle \lim_{x\to -5 } \frac{x^2+6x+5}{x+5} = \dots$

5. these are simple, but what about harder problems ( for example: sin, cos,... ) and other methods like hopital?

6. Originally Posted by danial
and other methods like hopital?
Says if $\displaystyle \lim_{x\to x_0} f(x) = \lim_{x\to x_0} g(x) = 0,\pm\infty$

then $\displaystyle \frac{\lim_{x\to x_0} f(x)}{\lim_{x\to x_0} g(x)} = \frac{\lim_{x\to x_0} f'(x)}{\lim_{x\to x_0} g'(x)}$

This one can be done by direct substitution, but try with L'hospitals instead, do you get the same result?

$\displaystyle \lim_{x\to 8 }\frac{x^2-64}{x-8}$

7. so, when can we use hoppital?

8. Originally Posted by danial
so, when can we use hoppital?
If by direct substitution you get $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$, then you can use L'Hospital's Rule.

The other indeterminate forms can be transformed to these forms using exponentials and logarithms.

9. I get the answer 16, is it right?

10. generally, how many methods there are to solve limits?

11. Too many to name. It takes a great deal of experience and ingenuity to know what to do in many different situations.

12. Is this true answer for this problem?

13. I get 16 for above problem, is it true?

14. One of the more ingenious methods of evaluating limits is using the Sandwich Theorem.

One of the most famous limits is evaluated this way: $\displaystyle \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} = 1$.

This requires using some knowledge of the unit circle.

The red length is $\displaystyle \cos{\theta}$, the green length is $\displaystyle \sin{\theta}$ and the purple length is $\displaystyle \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$.

Notice that the area of the sector is a little larger than that of the smaller triangle, and a little smaller than that of the larger triangle. As you make $\displaystyle \theta \to 0$, the three areas will end up becoming equal.

So $\displaystyle A_{\Delta_{\textrm{small}}} \leq A_{\textrm{Sector}} \leq A_{\Delta_{\textrm{large}}}$

$\displaystyle \frac{\sin{\theta}\cos{\theta}}{2} \leq \frac{r^2\theta}{2} \leq \frac{1\tan{\theta}}{2}$, and since $\displaystyle r=1$

$\displaystyle \frac{\sin{\theta}\cos{\theta}}{2} \leq \frac{\theta}{2} \leq \frac{\sin{\theta}}{2\cos{\theta}}$

$\displaystyle \lim_{\theta \to 0}\frac{\sin{\theta}\cos{\theta}}{2} \leq \lim_{\theta \to 0}\frac{\theta}{2} \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{2\cos{\theta}}$

$\displaystyle \lim_{\theta \to 0}\sin{\theta}\cos{\theta} \leq\lim_{\theta \to 0}\theta \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\cos{\theta}}$

$\displaystyle \lim_{\theta \to 0}\cos{\theta} \leq \lim_{\theta \to 0}\frac{\theta}{\sin{\theta}} \leq \lim_{\theta \to 0}\frac{1}{\cos{\theta}}$

$\displaystyle \lim_{\theta \to 0}\frac{1}{\cos{\theta}} \geq\lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \geq \lim_{\theta \to 0}\cos{\theta}$

$\displaystyle \lim_{\theta \to 0}\cos{\theta} \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \leq \lim_{\theta \to 0}\frac{1}{\cos{\theta}}$

$\displaystyle 1 \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \leq \frac{1}{1}$

$\displaystyle 1 \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \leq 1$.

It should be clear after being sandwiched that $\displaystyle \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} = 1$.

[Of course, to be pedantic, this only evaluates the right hand limit (i.e. making $\theta \to 0$ from the first quadrant where $\displaystyle \theta > 0$), but the left hand limit is evaluated in a near identical manner.]

15. Originally Posted by danial
I get 16 for above problem, is it true?
Yes, $\displaystyle \lim_{x \to 8}\frac{x^2-64}{x-8} = \lim_{x \to 8}\frac{(x-8)(x+8)}{x-8} = \lim_{x \to 8}(x+8) = 8+8 = 16$.

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