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Limit properties, resolving indeterminate forms, limits at infinity and continuity.
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Direct substitution gives the indeterminate form 0/0. Factor the difference of squares and cancel the common factor away from the approach point. Then calculate the limit of the simplified expression.
Do not subtract two infinities separately. Multiply by the conjugate, then divide numerator and denominator by x. Remember x is positive near +∞.
Determine the y value approached from the left and from the right separately. A two-sided limit exists only when they agree. The filled-point value does not by itself determine the limit.
Calculate the left-hand branch limit at the joining point. The right-hand branch value at that point must agree with it. Form an equation and solve for the parameter.
Divide numerator and denominator by their highest power of x. Terms containing x in the denominator tend to zero; evaluate the remaining constants.
Direct substitution gives 0/0. Multiply numerator and denominator by the conjugate of the numerator, using a plus sign. Apply the difference-of-squares identity, cancel the common factor and then substitute the approach value.
Apply the sum and constant-multiple rules for limits. Replace the functions by their limits, retaining coefficients and the subtraction sign.
Determine the sign of the difference between x and the approach point on the specified side. Remove the absolute-value signs accordingly, cancel and finish the calculation.
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https://www.catestia.com/en/mathematics/12th-grade/function-limits-and-continuity-test/200/