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# Need some more help finding limits

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I have a problem that I haven't learned how to solve yet. If someone could please provide the answer with steps that would be awesome.

$$\lim_{x\rightarrow 2}\frac{2x}{x^2-4}$$

May 6, 2018

#1
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Find the following limit:

lim_(x->2^-) (2 x)/(x^2 - 4)

If none of the terms of a product approach 0 as x->a, then by the product rule, the limit of the product is the product of the limits.

Applying the product rule, write lim_(x->2^-) (2 x)/(x^2 - 4) as 2 (lim_(x->2^-) x) (lim_(x->2^-) 1/(x^2 - 4)):

2 lim_(x->2^-) x lim_(x->2^-) 1/(x^2 - 4)

The limit of a continuous function at a point is just its value there.

lim_(x->2^-) x = 2:

2×2 lim_(x->2^-) 1/(x^2 - 4)

If lim_(x->2^-) y = 0, then the limit of 1/y is ± ∞.

Since lim_(x->2^-) (x^2 - 4) = 0 and x^2 - 4<0 for all x just to the left of x = 2, lim_(x->2^-) 1/(x^2 - 4) = -∞:

2×2 -∞

= -∞

May 6, 2018
#2
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\lim_{x\rightarrow 2}\frac{2x}{x^2-4}

$$\displaystyle\lim_{x\rightarrow 2}\;\frac{2x}{x^2-4}\\ \text{I will look at this from above and below}\\ Let \;\;\delta\;\;be\;a\;tiny\;positive\;number\;\\ \text{As x tends to 2 from above this will be} \;\; \frac{4}{\delta } =+\infty\\ \text{As x tends to 2 from below this will be} \;\; \frac{4}{-\delta } =-\infty \\$$ May 7, 2018