Limit as x goes to 5 from below

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In summary, when dividing by the highest power for $\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}$, the numerator goes to $1-4/5$ and the denominator goes to 0. This means that the limit does not exist. However, the left hand limit can still be evaluated, and it approaches negative infinity. This is because when dividing by a very small number (close to 0), the numerator becomes positive and the denominator becomes negative, resulting in a negative overall value.
  • #1
Guest2
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Dividing by the highest power for $\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}$ I get

$\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}= \lim_{x \to 5^{-}}\frac{1-4/x}{1/x^{99}-5/x^{100}}$

However the denominator goes to $0$ whereas the numerator goes to $1-4/5$

Why isn't dividing by the highest power working?
 
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  • #2
Guest said:
Dividing by the highest power for $\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}$ I get

$\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}= \lim_{x \to 5^{-}}\frac{1-4/x}{1/x^{99}-5/x^{100}}$

However the denominator goes to $0$ whereas the numerator goes to $1-4/5$

Why isn't dividing by the highest power working?

the reason in the given expression numerator is not zero but denominator is zero at x = 5.
hence after division also
 
  • #3
kaliprasad said:
the reason in the given expression numerator is not zero but denominator is zero at x = 5.
hence after division also
So how does one calculate the limit?
 
  • #4
Guest said:
So how does one calculate the limit?
\
so the limit does not exist as it goes to infinite
 
  • #5
Kali, you are correct that the limit does not exist, but the LEFT HAND limit can still be evaluated.

To the OP - what happens when you divide by a very, very small number (close to 0)?
 
  • #6
Prove It said:
Kali, you are correct that the limit does not exist, but the LEFT HAND limit can still be evaluated.

To the OP - what happens when you divide by a very, very small number (close to 0)?

you mean numerator is is positive and denominator is -ve( close to zero) so limit is - infinite ?
 
  • #7
kaliprasad said:
you mean numerator is is positive and denominator is -ve( close to zero) so limit is - infinite ?

That is correct :)
 

FAQ: Limit as x goes to 5 from below

What does "limit as x goes to 5 from below" mean?

When we say "limit as x goes to 5 from below", we are talking about the behavior of a function as the input value, x, approaches the value of 5 from values that are less than 5. In other words, we are looking at how the function behaves as x gets closer and closer to 5 from the left side of the number line.

How do you find the limit as x goes to 5 from below?

To find the limit as x goes to 5 from below, we evaluate the function at values of x that are getting closer and closer to 5 from the left side of the number line. We can do this by creating a table of values or by graphing the function. The limit will be the value that the function approaches as x gets closer and closer to 5 from below.

Why is it important to consider the limit as x goes to 5 from below?

The limit as x goes to 5 from below can give us insight into the behavior of a function at a specific point. It can also help us determine if a function is continuous at that point. Additionally, limits play a crucial role in calculus and other areas of mathematics.

What is the difference between the limit as x goes to 5 from below and the limit as x goes to 5 from above?

The limit as x goes to 5 from below is concerned with the behavior of a function as x approaches 5 from values that are less than 5. On the other hand, the limit as x goes to 5 from above looks at how the function behaves as x gets closer and closer to 5 from values that are greater than 5. These two limits may be different, indicating that the function has a discontinuity at x = 5.

Can the limit as x goes to 5 from below be undefined?

Yes, the limit as x goes to 5 from below can be undefined. This can occur if the function has a vertical asymptote or a jump discontinuity at x = 5. In these cases, the function does not approach a single value as x gets closer to 5 from below, so the limit is undefined.

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