Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$
$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$
✓
Answer
$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$ Dividing $\text{x}^4-3\text{x}^3+2\text{ by }\text{x}^3-5\text{x}^2+3\text{x}+1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ Dividing $\text{x}^3-5\text{x}^2+3\text{x}+1\text{ by }\text{x}-1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{(\text{x}-1)(\text{x}^2-4\text{x}-1)}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}}{\big(\text{x}^2-4\text{x}-1\big)}$ $=1+2+\frac{7}{(1-4-1)}$ $=3-\frac{7}{4}$ $=\frac{12-7}{4}$ $=\frac{5}{4}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ Dividing $\text{x}^3-5\text{x}^2+3\text{x}+1\text{ by }\text{x}-1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{(\text{x}-1)(\text{x}^2-4\text{x}-1)}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}}{\big(\text{x}^2-4\text{x}-1\big)}$ $=1+2+\frac{7}{(1-4-1)}$ $=3-\frac{7}{4}$ $=\frac{12-7}{4}$ $=\frac{5}{4}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A circle whose centre is the point of intersection of the lines 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0 passes through the origin. Find its equation.
If $\text{f(x)}=\frac{1}{1-\text{x}},$ show that $\text{f}\big[\text{f}\{\text{f(x)}\}\big]=\text{x}$
→Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\sqrt{\text{x}}-\sqrt{2}}$
→$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\sqrt{\text{x}}-\sqrt{2}}$
Sketch the graphs of the following trigonometric functions:
$\phi\text{(x)}=2\cos\Big(\text{x}-\frac{\pi}{6}\Big)$
→$\phi\text{(x)}=2\cos\Big(\text{x}-\frac{\pi}{6}\Big)$
Prove that $(2\sqrt{3}+3)\sin\text{x}+2\sqrt{3}\cos\text{x}$ lies between $-(2\sqrt{3}+\sqrt{15})$ and $(2\sqrt{3}+\sqrt{15}).$
→If $A$ and $B$ are mutually exclusive events such that $P(A) = 0.35$ and $P(B) = 0.45$, find:
→- $\text{P}(\text{A}\cup\text{B})$
- $\text{P}({\text{A}}\cap{\text{B}})$
- $\text{P}({\text{A}}\cap\overline{\text{B}})$
- $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas:
$y^2 + 4x + 4y - 3 = 0$
→$y^2 + 4x + 4y - 3 = 0$
Differentiate the following from first principle: $\frac{1}{\sqrt{3-\text{x}}}$
→Solve the following equations:
$2\sin^{2}\text{x}=3\cos\text{x},0\leq\text{x}\leq2\pi$
→$2\sin^{2}\text{x}=3\cos\text{x},0\leq\text{x}\leq2\pi$
The sides of a square are $x = 6, x = 9, y = 3$ and $y = 6.$ Find the equation of a circle drawn on the diagonal of the square as its diameter.
→