Hello
I want to learn all methods for solving limits
can someone help me?
Well here's a simple one, the method is called direct substitution.
If possible, just substitute the value into the function
$\displaystyle \displaystyle \lim_{x\to 0 } 3x+5 = 3(0)+5= 5$
Now you have a go
$\displaystyle \displaystyle \lim_{x\to 2 } 4x^2+2x-1 = \dots$
$\displaystyle \displaystyle \lim_{x\to 1 } \frac{x+3}{x^2+x-8} = \dots$
Then try this one, what happens?
$\displaystyle \displaystyle \lim_{x\to -5 } \frac{x^2+6x+5}{x+5} = \dots$
Says if $\displaystyle \displaystyle \lim_{x\to x_0} f(x) = \lim_{x\to x_0} g(x) = 0,\pm\infty $
then $\displaystyle \displaystyle \frac{\lim_{x\to x_0} f(x)}{\lim_{x\to x_0} g(x)} = \frac{\lim_{x\to x_0} f'(x)}{\lim_{x\to x_0} g'(x)}$
This one can be done by direct substitution, but try with L'hospitals instead, do you get the same result?
$\displaystyle \displaystyle \lim_{x\to 8 }\frac{x^2-64}{x-8}$
One of the more ingenious methods of evaluating limits is using the Sandwich Theorem.
One of the most famous limits is evaluated this way: $\displaystyle \displaystyle \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} = 1$.
This requires using some knowledge of the unit circle.
The red length is $\displaystyle \displaystyle \cos{\theta}$, the green length is $\displaystyle \displaystyle \sin{\theta}$ and the purple length is $\displaystyle \displaystyle \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$.
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 $\displaystyle \displaystyle \theta \to 0$, the three areas will end up becoming equal.
So $\displaystyle \displaystyle A_{\Delta_{\textrm{small}}} \leq A_{\textrm{Sector}} \leq A_{\Delta_{\textrm{large}}}$
$\displaystyle \displaystyle \frac{\sin{\theta}\cos{\theta}}{2} \leq \frac{r^2\theta}{2} \leq \frac{1\tan{\theta}}{2}$, and since $\displaystyle \displaystyle r=1$
$\displaystyle \displaystyle \frac{\sin{\theta}\cos{\theta}}{2} \leq \frac{\theta}{2} \leq \frac{\sin{\theta}}{2\cos{\theta}}$
$\displaystyle \displaystyle \lim_{\theta \to 0}\frac{\sin{\theta}\cos{\theta}}{2} \leq \lim_{\theta \to 0}\frac{\theta}{2} \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{2\cos{\theta}}$
$\displaystyle \displaystyle \lim_{\theta \to 0}\sin{\theta}\cos{\theta} \leq\lim_{\theta \to 0}\theta \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\cos{\theta}}$
$\displaystyle \displaystyle \lim_{\theta \to 0}\cos{\theta} \leq \lim_{\theta \to 0}\frac{\theta}{\sin{\theta}} \leq \lim_{\theta \to 0}\frac{1}{\cos{\theta}}$
$\displaystyle \displaystyle \lim_{\theta \to 0}\frac{1}{\cos{\theta}} \geq\lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \geq \lim_{\theta \to 0}\cos{\theta}$
$\displaystyle \displaystyle \lim_{\theta \to 0}\cos{\theta} \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \leq \lim_{\theta \to 0}\frac{1}{\cos{\theta}}$
$\displaystyle \displaystyle 1 \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \leq \frac{1}{1}$
$\displaystyle \displaystyle 1 \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \leq 1$.
It should be clear after being sandwiched that $\displaystyle \displaystyle \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} = 1$.
[Of course, to be pedantic, this only evaluates the right hand limit (i.e. making $\displaystyle \theta \to 0$ from the first quadrant where $\displaystyle \displaystyle \theta > 0$), but the left hand limit is evaluated in a near identical manner.]