Question 1. Evaluate \(\displaystyle \lim_{x\to 1}\frac{x^4-1}{x-1}\).
Factor the numerator using the difference of powers: \(x^4-1=(x-1)(x^3+x^2+x+1)\). For \(x\ne 1\), cancel \(x-1\): \(\displaystyle \frac{x^4-1}{x-1}=x^3+x^2+x+1\). Now substitute \(x=1\): \(1^3+1^2+1+1=4\). Answer: C. \(4\)
Question 2. Evaluate \(\displaystyle \lim_{x\to 0}\frac{x^3+12x^2-5x}{5x}\).
Factor \(x\) from the numerator: \(x^3+12x^2-5x=x(x^2+12x-5)\). For \(x\ne 0\), cancel \(x\): \(\displaystyle \frac{x(x^2+12x-5)}{5x}=\frac{x^2+12x-5}{5}\). Substitute \(x=0\): \(\displaystyle \frac{0^2+12(0)-5}{5}=-1\). Answer: D. \(-1\)
Question 3. Evaluate \(\displaystyle \lim_{x\to 3}\frac{x^2+6x-27}{x-3}\).
Factor the numerator: \(x^2+6x-27=(x-3)(x+9)\). For \(x\ne 3\), cancel \(x-3\): \(\displaystyle \frac{(x-3)(x+9)}{x-3}=x+9\). Substitute \(x=3\): \(3+9=12\). Answer: C. \(12\)
Question 4. Evaluate \(\displaystyle \lim_{h\to 0}\frac{(x+h)^3-x^3}{h}\).
Expand the cube: \((x+h)^3=x^3+3x^2h+3xh^2+h^3\). Subtract \(x^3\): \((x+h)^3-x^3=3x^2h+3xh^2+h^3\). Factor and cancel \(h\): \(\displaystyle \frac{h(3x^2+3xh+h^2)}{h}=3x^2+3xh+h^2\). Let \(h\to 0\): \(3x^2+3x(0)+0^2=3x^2\). Answer: D. \(3x^2\)
Question 5. Evaluate \(\displaystyle \lim_{x\to -6}\frac{x^2+12x+36}{x+6}\).
Factor the numerator: \(x^2+12x+36=(x+6)^2\). For \(x\ne -6\), cancel one factor of \(x+6\): \(\displaystyle \frac{(x+6)^2}{x+6}=x+6\). Substitute \(x=-6\): \(-6+6=0\). Answer: A. \(0\)
Question 6. Evaluate \(\displaystyle \lim_{x\to -2}\frac{1}{x+2}\).
Examine the one-sided limits. As \(x\to -2^{-}\), \(x+2\) is a very small negative number, so \(\displaystyle \frac{1}{x+2}\to -\infty\). As \(x\to -2^{+}\), \(x+2\) is a very small positive number, so \(\displaystyle \frac{1}{x+2}\to \infty\). Because the one-sided limits are not equal, the two-sided limit does not exist. Answer: A. Does not exist
Question 7. Evaluate \(\displaystyle \lim_{x\to -1^{-}}\frac{1}{x+1}\).
As \(x\to -1^{-}\), the values of \(x\) are slightly less than \(-1\). Therefore, \(x+1\) is negative and approaches \(0\). The reciprocal of a negative number approaching \(0\) decreases without bound: \(\displaystyle \frac{1}{x+1}\to -\infty\). Answer: A. \(-\infty\)
Question 8. Evaluate \(\displaystyle \lim_{x\to 6^{-}}\frac{1}{(x-6)^2}\).
As \(x\to 6^{-}\), \(x-6\) approaches \(0\) through negative values. Squaring makes the denominator positive: \((x-6)^2\to 0^{+}\). The reciprocal of a positive number approaching \(0\) increases without bound: \(\displaystyle \frac{1}{(x-6)^2}\to \infty\). Answer: D. \(\infty\)
Question 9. Evaluate \(\displaystyle \lim_{x\to -\infty}\frac{5}{5-\left(\frac{3}{x^2}\right)}\).
As \(x\to -\infty\), \(x^2\to \infty\), so \(\displaystyle \frac{3}{x^2}\to 0\). Substitute this limiting value into the denominator: \(\displaystyle 5-\frac{3}{x^2}\to 5-0=5\). Therefore, \(\displaystyle \lim_{x\to -\infty}\frac{5}{5-\left(\frac{3}{x^2}\right)}=\frac{5}{5}=1\). Answer: C. \(1\)
Question 10. Evaluate \(\displaystyle \lim_{x\to \infty}\frac{5x+1}{11x-7}\).
Divide the numerator and denominator by \(x\): \(\displaystyle \frac{5x+1}{11x-7}=\frac{5+\frac{1}{x}}{11-\frac{7}{x}}\). As \(x\to \infty\), both \(\frac{1}{x}\to 0\) and \(\frac{7}{x}\to 0\). Thus the limit is \(\displaystyle \frac{5+0}{11-0}=\frac{5}{11}\). Answer: B. \(\frac{5}{11}\)
