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How to calculate $\lim\limits_{x\to5} 2\cdot \frac {x+2} {x-5}$ without L'Hopital?

I'm making a lot of progress in learning how to calculate limits, but some still seem to be impossible for me. I don't want a concrete solution but if someone could provide a hint or a trick to progress in general it would be really helpful. I'm having a hard time here.

6 Answers 6

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Direct substitution gives

$$\lim_{x\to5}2\cdot\frac{x+2}{x-5}=2\cdot\frac70\to\pm\infty$$

Can you check this reasoning?

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    I understand that $\frac 1 0 \to \infty$, but on other examples I was able to cancel the $0$ from the divisor with some way. How can I be sure that there is no way for this example? – 2017-01-24
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    @user405981 Well, to cancel the $0$, you need a $0$ in the numerator. There is none. So to prove this diverges... well, do you have notes on how to do that? – 2017-01-24
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    - Ahhh okay I have an idea now how to solve it. Thank you. – 2017-01-24
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    :-) Ok. Hope it works, and if not, well... I guess we'll be here. – 2017-01-24
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The limit in the title does not exist and hence the calculation is surely "impossible".

Note that $$ \lim_{x\to 5+}\frac{2(x+2)}{x-5}=\infty,\quad \lim_{x\to 5-}\frac{2(x+2)}{x-5}=-\infty. $$

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    The tag `limits-without-lhopital` is meaningless for this particular question. Moreover, the question seems suggesting that the limit *exists* in the first place, but no, it does not exists, not even in the sense of extended real numbers. One could say that $\lim_{x\to 0+}\frac{1}{x}=\infty$ but the limit $\lim_{x\to 0}\frac{1}{x}$ does not exist at all even in the extended real number system. – 2017-01-24
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In this case, you can prove that, given $M > 0$, exists $x \in \mathbb{R} \setminus \{5\}$ such that $|f(x)| \geq M$, where $$ f : x \mapsto \frac{14}{x - 5} $$ and it proves that $|f(x)| \longrightarrow \infty$ when $x \longrightarrow 5$, because $$ f(x) + 2 = 2 \frac{x + 2}{x - 5}\mbox{.} $$

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  1. $\lim_{x \to 5} \frac{2*(x+2)}{x-5}$ , divide numerator and
    denominator by $x$ then you have
    $\frac{2+\frac{2}{x}}{1-\frac{5}{x}}$. Now plug in the value of $x$ and you wil get something over $ 0$ and it is not possible to calculate this i think. Hint: while calculating limits $==> $ try to bring first term in some order, like in this example i divided both(numerator and denminator) by $x$. In most cases first term gives you better idea to calculate limit.
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A different approach.. $$ \frac{x+2}{x-5} = \frac{x-5 + 7}{x-5} = 1 + \frac{7}{x-5} $$ So $$ \lim_{x\to 5}\frac{x+2}{x-5} =\lim_{x\to 5} 1 + \frac{7}{x-5} = 1 +\lim_{x\to 5} \frac{7}{x-5} $$ (if this makes more sense as we only varying the denominator)

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First of all that $"2"$ is rather redundant. Let's put it aside and focus on $$l=\lim_{x\rightarrow5}\frac{x+2}{x-5}$$

Let $$u=x-5$$

We have $$l=\lim_{u\rightarrow0}\frac{u+7}{u}=\lim_{u\rightarrow0}\frac{u}{u}+\lim_{u\rightarrow0}\frac{7}{u}=1+7\lim_{u\rightarrow0}\frac{1}{u}$$

Now, what can you tell about how the function $g(u)=\frac{1}{u}$ "behaves" as $u\rightarrow0$ ?

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    Could the down-voter explain his reasoning? – 2017-01-24