Question

Can anyone help me understand how to solve Limit problems 1, 2, 4, 6, 8, 9? I have my practice sheet attached below.

Answer 1

1 and 2 were answered here: brainly.com/question/13308478

4. Factor the denominator as a sum of cubes:

$x^3+8=(x+2)(x^2-2x+4)$

Then

$\displaystyle\lim_{x\to-2}\frac{x+2}{x^3+8}=\lim_{x\to-2}\frac1{x^2-2x+4}=\frac1{12}$

6. Multiply the numerator and denominator by the conjugate of the numerator. This gives a difference of squares in the numerator that doesn't involve any square roots:

$\dfrac{3-\sqrt{x+5}}{x-4}\dfrac{3+\sqrt{x+5}}{3+\sqrt{x+5}}=\dfrac{9-(x+5)}{(x-4)(3+\sqrt{x+5})}$

Then $9-(x+5)=-(x-4)$ which cancels with the factor of $x-4$ in the denominator:

$\displaystyle\lim_{x\to4}\frac{3-\sqrt{x+5}}{x-4}=-\lim_{x\to4}\frac1{3+\sqrt{x+5}}=-\frac16$

8. Combine the fractions in the numerator:

$\dfrac1{2+x}-\dfrac12=\dfrac2{2(2+x)}-\dfrac{2+x}{2(2+x)}=-\dfrac x{2(2+x)}$

Then

$\displaystyle\lim_{x\to0}\frac{\frac1{2+x}-\frac12}x=-\lim_{x\to0}\frac x{2x(2+x)}=-\lim_{x\to0}\frac1{2(2+x)}=-\frac14$

9. Factor 5 out of the denominator:

$x-10=5\left(\dfrac x5-2\right)$

so that

$\displaystyle\lim_{x\to10}\frac{\frac x5-2}{x-10}=\lim_{x\to10}\frac15=\frac15$