Limit approaching 2a

Question

When I substitute 2a into x, I get 0 but the answer should be 2.. Any help will be appreciated!

Answer 1

HINT:

$x^2+ax-6a^2=(x-2a)(x+3a)$

$5x^2-10ax=5x(x-2a)$

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$$\frac{5x^2-10ax}{x^2+ax-6a^2}=\frac{5x(x-2a)}{(x-2a)(x+3a)}=\frac{5x}{(x+3a)}$$

we get,

$$\lim_{x\to 2a}\frac{5x^2-10ax}{x^2+ax-6a^2}=\lim_{x\to 2a}\frac{5x}{(x+3a)}=\frac{10a}{5a}=2$$