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I need to find $$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$

Looking at the graph, I know the answer should be $\frac{20}{17}$, but when I tried solving it, I reached $0$.

Here are the two ways I approached this:

WAY I:

$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} = \lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{x^2}(2- \frac{50}{x^2})} {\require{cancel} \cancel{x^2}(2+ \frac{3}{x}-\frac{35}{x^2})} =\frac{2-2}{\frac {42}{5}}=0 $$

WAY II: $$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} = \lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{2}(x^2- 25)} {\require{cancel} \cancel{2}(x^2+ \frac{3}{2}x-\frac{35}{2})} =\lim_{x\rightarrow -5} \frac{{\require{cancel} \cancel{(x-5)}}(x+5)}{{\require{cancel} \cancel{(x-5)}}(x+3.5)}= \frac{-5+5}{-5+3.5}=0 $$

What am I doing wrong here?

Thanks!

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  • $\begingroup$ Generally $0$ comes when we assume something or enforce partial limits. $\endgroup$
    – Akash Roy
    Commented Nov 22, 2018 at 17:03
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    $\begingroup$ On multiplying the denominators you will get $-3\over2$ not $+3\over2$ as coefficient of x. $\endgroup$
    – Love Invariants
    Commented Nov 22, 2018 at 17:09
  • $\begingroup$ @LoveInvariants you mean on step 2 of way 2? the factorization? $\endgroup$
    – Netanel
    Commented Nov 22, 2018 at 17:17

4 Answers 4

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Numerator

$2x^2-50=2(x-5)(x+5)$.

Denominator

$2x^2+3x -35 =(2x-7)(x+ 5)$

$\dfrac{2(x-5)(x+5)}{(2x-7)(x+5)}=$

$\dfrac{2(x-5)}{2x+7}.$

Take the limit $x \rightarrow -5.$

Try to factorize the original expression.The term $(x+5) $ cancels out .

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  • $\begingroup$ Thank you. Why is my way of factorizing wrong though? $\endgroup$
    – Netanel
    Commented Nov 22, 2018 at 17:19
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    $\begingroup$ Netanel. The term (x+5) goes to zero,factor in numerator and denominator, cancels .I do not see it canceling in your calculation, so taking the limit numerator and denominator should go to zero .Please check .Denominator in Way1 =0!! $\endgroup$
    – Peter Szilas
    Commented Nov 22, 2018 at 17:27
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For your way $1$, check the computation of your denominator, it should give you $0$ again.

For your way $2$, check your factorization in your denominator as well.

Use L'hopital's rule:

$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}= \lim_{x\rightarrow -5} \frac{4x}{4x+3}=\frac{-20}{-17}=\frac{20}{17}$$

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  • $\begingroup$ Thank you. Can you point out the mistake in Way 2? $\endgroup$
    – Netanel
    Commented Nov 22, 2018 at 17:17
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    $\begingroup$ $(x-5)(x+3.5)=x^2\color{red}-1.5x -\frac{35}2$. That is the factorization is not correct. $\endgroup$
    – Siong Thye Goh
    Commented Nov 22, 2018 at 17:19
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Hint: Try factorization!

$$ \frac{2x^2-50}{2x^2+3x-35}=\frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=\frac{4(x-5)(x+5)}{(2x+10)(2x-7)} $$

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As an alternative by $y=x+5 \to 0$

$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}=\lim_{y\rightarrow 0} \frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=\lim_{y\rightarrow 0} \frac{2y^2-20y}{2y^2-17y}=\lim_{y\rightarrow 0} \frac{2y-20}{2y-17}$$

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  • $\begingroup$ Thank you. Why is my way wrong though? $\endgroup$
    – Netanel
    Commented Nov 22, 2018 at 17:18
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    $\begingroup$ @Netanel In the first on we have $$2+ \frac{3}{x}-\frac{35}{x^2}=2- \frac{3}{5}-\frac{35}{25}=\frac{50-15-35}{25}=0$$ $\endgroup$
    – user
    Commented Nov 22, 2018 at 17:20
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    $\begingroup$ For the second one $$x^2+ \frac{3}{2}x-\frac{35}{2}=\frac12(2x-7)(x+5)$$ $\endgroup$
    – user
    Commented Nov 22, 2018 at 17:40

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