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let $f(x) = 4x-13$

How do I simplify the limit below to find if it exists? $$\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$$

The limit involves two variables and I'm unable to remove $h$ from the denumerator.

I first tried replacing the function with its content $$\lim_{h\to0}\dfrac{4(x+h)-13-4x-13}{h}$$ I can do this small simplification $$\lim_{h\to0}\dfrac{4(x+h)-4x-26}{h}$$ However, I do not know where to go from here

Cedric Martens
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Note that you made mistake here $$\lim_{h\to0}\dfrac{4(x+h)-13-4x-\color{red}{13}}{h}$$ it should be $$\lim_{h\to0}\dfrac{4(x+h)-13-4x\color{red}{+13}}{h}$$

Given $f(x)=4x-13$ $$\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$$ $$\lim_{h\to0}\frac{4(x+h)-13-4x+13}{h}$$ $$\lim_{h\to0}\frac{4x+4h-4x}{h}$$ $$\lim_{h\to0}\frac{4h}{h}$$ $$=4$$

Key Flex
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  • Thanks for the answer! I don't understand why $$\lim_{h\to0}\dfrac{4h}{h}=4$$ If you replace $h$ by 0. You should get $$\dfrac{0}{0}$$ no? – Cedric Martens Jun 10 '18 at 17:23
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    $h$ in the numerator and in the denominator gets cancelled and you will be left with only $4$ – Key Flex Jun 10 '18 at 17:25
  • @CedricMartens: See this question: https://math.stackexchange.com/questions/2628911/why-are-we-allowed-to-cancel-fractions-in-limits. (L'Hospital's rule is certainly not the “right” explanation here, in my opinion.) – Hans Lundmark Jun 10 '18 at 19:07