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Math Help - limits

  1. #1
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    limits

    Hello
    I want to learn all methods for solving limits
    can someone help me?
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  2. #2
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  3. #3
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    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
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  4. #4
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    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
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  5. #5
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    these are simple, but what about harder problems ( for example: sin, cos,... ) and other methods like hopital?
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  6. #6
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    Quote Originally Posted by danial View Post
    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}
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  7. #7
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    so, when can we use hoppital?
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  8. #8
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    Quote Originally Posted by danial View Post
    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.
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  9. #9
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    I get the answer 16, is it right?
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  10. #10
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    generally, how many methods there are to solve limits?
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  11. #11
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    Too many to name. It takes a great deal of experience and ingenuity to know what to do in many different situations.
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  12. #12
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    Is this true answer for this problem?
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  13. #13
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    I get 16 for above problem, is it true?
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  14. #14
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    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.]
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  15. #15
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    Quote Originally Posted by danial View Post
    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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