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Section 13.1 : Limits
2. Evaluate the following limit.
\[\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,1} \right)} \frac{{{x^2} - 2xy}}{{{x^2} - 4{y^2}}}\]Show All Steps Hide All Steps
Start SolutionOkay, with this problem we can see that, if we plug in the point, we get zero in the numerator and the denominator. Unlike most of the examples of this type however, that doesn’t just mean that the limit won’t exist.
In this case notice that we can factor and simplify the function as follows,
\[\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,1} \right)} \frac{{{x^2} - 2xy}}{{{x^2} - 4{y^2}}} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,1} \right)} \frac{{x\left( {x - 2y} \right)}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,1} \right)} \frac{x}{{x + 2y}}\]We may not be used to factoring these kinds of polynomials but we can’t forget that factoring is still a possibility that we need to address for these limits.
Show Step 2Now, that we’ve factored and simplified the function we can see that we’ve lost the division by zero issue and so we can now evaluate the limit. Doing this gives,
\[\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,1} \right)} \frac{{{x^2} - 2xy}}{{{x^2} - 4{y^2}}} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,1} \right)} \frac{x}{{x + 2y}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}}}\]