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$$ \text{Given}\; \lim_{x \to 1} \frac{f(x)-4}{x-1} = 10, \;\text{evaluate}\; \lim_{x \to 1} f(x) $$

I'm wondering if anyone can give me some tips on how to approach this problem. I don't even know where to start.

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  • $\begingroup$ HINT: $\frac{\infty}{\infty}$ $\endgroup$
    – Simple
    Aug 28, 2015 at 23:49
  • $\begingroup$ @adam19325 Just for future reference, please include all your working next time in your question, even if you haven't been able to do much, only so that we can see where a problem may lie and better help you. $\endgroup$ Aug 29, 2015 at 0:18

3 Answers 3

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Since it appears that you're doing basic calculus, I'll put my answer in terms of that.

The definition for the slope of a tangent line on a curve using limits is as follows

$$ \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = m $$

Now let's look at what they give you,

$$ \lim_{x \to 1} \frac{f(x)-4}{x-1} = 10 $$

Finally, let's look at what they ask of you,

$$ \lim_{x \to 1} f(x) = ? $$

If you look at the definition and what they've given you, you'll see that they have given you $f(1)$. Since the function is continuous at $x=1$,

$$\lim_{x \to 1} f(x) = f(1)$$

Therefore,

$$\lim_{x \to 1} f(x) = 4$$

Always remember to look for any correlation between your equations and your problem if you ever get stuck.

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  • $\begingroup$ Explained very well for the OPs level of understanding. +1 $\endgroup$ Aug 29, 2015 at 0:15
  • $\begingroup$ where is it given that $f$ is continuous and also that $f(1) = 4$. $\endgroup$
    – Paramanand Singh
    Aug 30, 2015 at 3:37
  • $\begingroup$ @ParamanandSingh I didn't say that $f$ was continuous, I said that $f$ was continuous at $x=1$. This is given by the presence of a defined slope at $x=1$, which could not exist if $f$ was not continuous at $x=1$. $\endgroup$
    – Slinky
    Aug 30, 2015 at 3:39
  • $\begingroup$ How do you know that $f$ is continuous at $x = 1$? Also where is it provided that $f(1) = 4$? A proper answer is the one I have provided. $\endgroup$
    – Paramanand Singh
    Aug 30, 2015 at 3:41
  • $\begingroup$ @ParamanandSingh "This is given by the presence of a defined slope at $x=1$, which could not exist if $f$ was not continuous at $x=1$." $f(1)$ is provided by correlating the definition with the provided equation. $\endgroup$
    – Slinky
    Aug 30, 2015 at 3:47
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By illegal calculations, we get $\lim_{x \to 1}f(x) = 4$. To prove that now, use the $\epsilon - \delta$ definition of limits, with $\delta = \epsilon/10$.

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Use the algebra of limits to get $$\lim_{x \to 1}f(x) = \lim_{x \to 1}f(x) - 4 + 4 = \lim_{x \to 1}\frac{f(x) - 4}{x - 1}\cdot (x - 1) + 4 = 10\cdot 0 + 4 = 4$$

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