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so I just got through my first Calculus lesson in class, introduction to limits.

One of the questions was:

$\displaystyle \lim_{x \to 1} \frac{\frac 1x-1}{x-1}$

When you substitute 1 into the function, it equals $\frac 00$. My teacher said that when you get $\frac 00$, you should either factor, rationalize the denom/numerator, or multiply by the conjugate to manipulate the equation. I've tried those and it hasn't worked. Any help please?

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Hint:\begin{align} \lim_{x \to 1} \frac{\frac 1x-1}{x-1}&=\lim_{x \to 1} \frac{\frac 1x-1}{x-1} \frac xx \\ &=\lim_{x \to 1} \frac{1-x}{x(x-1)} \\ &=\lim_{x \to 1} \frac{-(x-1)}{x(x-1)} \\ &= \cdots \end{align} Cancel the $(x-1)$'s and try to evaluate the limit from there.

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  • $\begingroup$ So it's -1! Wow thanks, I didn't see that. So that means if you have a fraction in the numerator, a good method would be to try to cancel it out, which you did by multiplying the top and bottom by x. $\endgroup$ – NoobCoder Feb 8 '15 at 2:02
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    $\begingroup$ Exactly. We generally do not like fractions within a larger fraction. $\endgroup$ – Cookie Feb 8 '15 at 2:02
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Hope this blurb helps:

$$\lim_{x\to1} \frac{\frac{1}{x}-1}{x-1}$$

This simplifies to:

$$\lim_{x\to1} \frac{\frac{1-x}{x}}{x-1}$$

Multiplying out using fraction rules:

$$\lim_{x\to1} {\frac{1-x}{x}}*\frac{1}{(x-1)}$$

and to cancel terms:

$$\lim_{x\to1} {\frac{1-x}{x}}*\frac{1}{-(1-x)}$$

Simplifying to get:

$$\lim_{x\to1} \frac{-1}{x} = -1$$

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  • $\begingroup$ The limit as $x \to 1$ is actually $-1$. $\endgroup$ – Cookie Feb 8 '15 at 2:00
  • $\begingroup$ Yeah just changed it, saw it myself right after I posted it. $\endgroup$ – Victor Feb 8 '15 at 2:02
  • $\begingroup$ Another good way to approach it. Thanks! I did your early steps, but factoring out the negative in the third step didn't occur to me. $\endgroup$ – NoobCoder Feb 8 '15 at 2:03
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Another method is to interpret this limit as the definition of the derivative of $f(x) = \dfrac{1}{x}$ at $x=1$.

Recall that:

$$f'(c) = \lim_{x \to c} \dfrac{f(x)-f(c)}{x-c}.$$

Then,

$$ \lim_{x \to 1} \dfrac{\frac{1}{x} - 1}{x-1} = f'(1).$$

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    $\begingroup$ I appreciate the response, but as I indicated above, I have only learned limits so far :). $\endgroup$ – NoobCoder Feb 8 '15 at 2:06
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    $\begingroup$ @NoobCoder A way to think about it related to this then is, the numerator around 1 is approaching 0 and is decreasing at "right 1 down 1" while the denominator is approaching 0 and increasing at "right 1 up 1". So when comparing the rate of change of them, dividing the -1 by 1, you get -1 – near 1 the rate of change is what matters. Basically that is Lhopital's rule and how linearization approximates a function near a point, which you'll learn later. $\endgroup$ – bjb568 Feb 8 '15 at 4:57
  • $\begingroup$ @NoobCoder Sorry, I didn't see that you haven't learned about derivatives yet. Hopefully this will inspire you to think about some future concepts :) $\endgroup$ – MathMajor Feb 8 '15 at 5:59

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