Limits

rachelmaddie said:
Hi. I need this checked please. LLPView attachment 20993
Click to expand...

rachelmaddie, your method would be to approximate the limit.
You have errors. 4.97/-0.02 is not equal to, nor is it approximately
equal to, 3.5. Also, 3.5 is not the correct value for your approximation.

For x = 4.99, the numerator is -0.0699 exactly and the denominator is
-0.02 as you have shown it. Therefore, the value of that fraction is 3.495.
Done in a similar way for x = 5.01, the value of that fraction is 3.505.

From this work, you don't know that the limit is 3.5 as x approaches 5.
It appears so.

Instead, look at factoring the parts of the algebraic fraction and reduce
before substituting x = 5 to evaluate the limit. Using the binomial factor
from the denominator as I gave here can often make it easier to factor
the expression in the numerator.


\(\displaystyle \dfrac{( \ \ \ \ \ \ \ \ \ )( \ \ \ \ \ \ \ \ \ )}{2(x - 5)}\)
 
Last edited:
lookagain said:
rachelmaddie, your method would be to approximate the limit.
You have errors. 4.97/-0.02 is not equal to, nor is it approximately
equal to, 3.5. Also, 3.5 is not the correct value for your approximation.

For x = 4.99, the numerator is -0.0699 exactly and the denominator is
-0.02 as you have shown it. Therefore, the value of that fraction is 3.495.
Done in a similar way for x = 5.01, the value of that fraction is 3.505.

From this work, you don't know that the limit is 3.5 as x approaches 5.
It appears so.

Instead, look at factoring the parts of the algebraic fraction and reduce
before substituting x = 5 to evaluate the limit. Using the binomial factor
from the denominator as I gave here can often make it easier to factor
the expression in the numerator.


\(\displaystyle \dfrac{( \ \ \ \ \ \ \ \ \ )( \ \ \ \ \ \ \ \ \ )}{2(x - 5)}\)
Click to expand...
553880FE-E778-4001-A07C-669FCA38D62B.jpeg
 
You did not tell us that you already knew that method! Why did you try the other method instead?
 
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