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Is it that all there is to evaluating limits algebraically: just substitute the number $\mathrm{x}$ is approaching in the given expression?
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Not always, but this often does happen, and when it does, we say that the function is continuous at the value of $x$ in question.

Look at this example:

$$\lim _{x \rightarrow 5} \frac{\sqrt{5} x-3}{x-3}$$

Substituting $x=5 \text { gives } \dfrac{\sqrt{25}-3}{5-3}=1$
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