Evaluate the lim x -> 2^+ (x^2 + 3x - 10)/(x^2 - 4x + 4)

This question looks difficult, but it is too easy once you learn to factorize the polynomials.

If we substitute the value of x = 2 in the expression we'll get 0/0. Sometimes, that can be the answer, but in this case, we can factorize it, take common, and transform those expressions into a simpler form. So let's do that.

$lim_{x \, \to \, 2^+} \frac{x^2 + 3x - 10}{x^2 - 4x + 4}$

= $lim_{x \, \to \, 2^+} \frac{(x-2)(x+5)}{(x-2)^2}$

If you are confused,
= x² + 3x - 10
= x² + 5x - 2x - 10
= x(x + 5) - 2(x + 5)
= (x−2)(x+5)

And for another expression, I used (a-b)² = a² + 2ab + b² formula. Let's get moving!

= $lim_{x \, \to \, 2^+} \frac{(x+5)}{(x-2)}$

If we substitute the value of x = 2, we'll get the number/0 result which means infinity. Now we need to find if it is positive or negative infinity.

The value of x approaches 2 from the right side (because there is + sign in the superscript of 2), so let's assume the value of x to be 2.001, 2.0001, ...

Now we can figure out that the numerator is +ve and the denominator is also +ve. So our result is +ve infinity.

= $lim_{x \, \to \, 2^+} \frac{(x+5) \, \rarr \, +ve}{(x-2) \, \rarr \, +ve}$ = +∞