GO CLASSES DPP | Calculus | Limits | GATE CE

Answer 1

\[
\lim_{x \to 0} \frac{x^3 + x^2}{2x^3 - 7x^2}
= \lim_{x \to 0} \frac{x^2(x + 1)}{x^2(2x - 7)}
= \lim_{x \to 0} \frac{x + 1}{2x - 7}
= \frac{1}{-7}
= \boxed{-\frac{1}{7}}
\]
 

Answer 2

Option B)  Step1 -  Take x^2 common from both num and denom and cancel out

                 Step2 -  You will get to know that denominator is not 0 ! Put the value

Answer 3

$$
f(x)=\lim _{x \rightarrow 0}\left[\frac{x^3+x^2}{2 x^3-7 x^2}\right]
$$

Since this has $\frac{0}{0}$ form, limit can be found by repeated application of L'Hospitals rule.

$$
\begin{aligned}
f(x) & =\lim _{x \rightarrow 0}\left[\frac{3 x^2+2 x}{6 x^2-14 x}\right] \\
& =\lim _{x \rightarrow 0}\left[\frac{6 x+2}{12 x-14}\right] \\
& =\left[\frac{6 \times 0+2}{12 \times 0-14}\right] \\
& =-\frac{1}{7}
\end{aligned}
$$