Evaluate the following limits:

\displaystyle \textbf{Question 1: }\lim\limits_{x\to1}\frac{x^2+1}{x+1}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to1}\frac{x^2+1}{x+1}=\frac{1^2+1}{1+1}=\frac{2}{2}=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\lim\limits_{x\to0}\frac{2x^2+3x+4}{x^2+3x+2}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to0}\frac{2x^2+3x+4}{x^2+3x+2}=\frac{2(0)^2+3(0)+4}{(0)^2+3(0)+2}=\frac{4}{2}=2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\lim\limits_{x\to3}\frac{\sqrt{2x+3}}{x+3}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to3}\frac{\sqrt{2x+3}}{x+3}=\frac{\sqrt{2(3)+3}}{3+3}=\frac{3}{6}=\frac12.
\displaystyle \\

\displaystyle \textbf{Question 4: }\lim\limits_{x\to1}\frac{\sqrt{x+8}}{\sqrt{x}}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to1}\frac{\sqrt{x+8}}{\sqrt{x}}=\frac{\sqrt{1+8}}{\sqrt1}=3.
\displaystyle \\

\displaystyle \textbf{Question 5: }\lim\limits_{x\to a}\frac{\sqrt{x}+\sqrt{a}}{x+a}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to a}\frac{\sqrt{x}+\sqrt{a}}{x+a}=\frac{\sqrt{a}+\sqrt{a}}{a+a}=\frac{2\sqrt{a}}{2a}=\frac{1}{\sqrt{a}}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\lim\limits_{x\to1}\frac{1+(x-1)^2}{1+x^2}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to1}\frac{1+(x-1)^2}{1+x^2}=\frac{1+(1-1)^2}{1+1^2}=\frac12.
\displaystyle \\

\displaystyle \textbf{Question 7: }\lim\limits_{x\to0}\frac{x^{2/3}-9}{x-27}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to0}\frac{x^{2/3}-9}{x-27}=\frac{0^{2/3}-9}{0-27}=\frac{0-9}{-27}=\frac13.
\displaystyle \\

\displaystyle \textbf{Question 8: }\lim\limits_{x\to0}9.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to0}9=9.
\displaystyle \text{Since }f(x)=9\text{ is a constant function, its limit is }9.
\displaystyle \\

\displaystyle \textbf{Question 9: }\lim\limits_{x\to2}(3-x).
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to2}(3-x)=3-2=1.
\displaystyle \\

\displaystyle \textbf{Question 10: }\lim\limits_{x\to-1}(4x^2+2).
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to-1}(4x^2+2)=4(-1)^2+2=4+2=6.
\displaystyle \\

\displaystyle \textbf{Question 11: }\lim\limits_{x\to-1}\frac{x^3-3x+1}{x-1}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to-1}\frac{x^3-3x+1}{x-1}=\frac{(-1)^3-3(-1)+1}{-1-1}=\frac{3}{-2}=-\frac32.
\displaystyle \\

\displaystyle \textbf{Question 12: }\lim\limits_{x\to0}\frac{3x+1}{x+3}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to0}\frac{3x+1}{x+3}=\frac{3(0)+1}{0+3}=\frac13.
\displaystyle \\

\displaystyle \textbf{Question 13: }\lim\limits_{x\to3}\frac{x^2-9}{x+2}.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to3}\frac{x^2-9}{x+2}=\frac{3^2-9}{3+2}=\frac{9-9}{5}=0.
\displaystyle \\

\displaystyle \textbf{Question 14: }\lim\limits_{x\to0}\frac{ax+b}{cx+d},\ d\neq0.
\displaystyle \text{Answer:}
\displaystyle \lim\limits_{x\to0}\frac{ax+b}{cx+d}=\frac{a(0)+b}{c(0)+d}=\frac{b}{d}.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.