Question 13 Marks
Find the value of $\lim _{x \rightarrow 0} \frac{\sqrt[n]{x+1}-1}{x}$.
View full question & answer→Questions
3 Marks Each
🎯
Test yourself on this topic
43 questions · timed · auto-graded
Question 23 Marks
Find the value of $\lim _{x \rightarrow 3} \frac{x^{5}-243}{x^{3}-27}$.
View full question & answer→Question 33 Marks
Find the value of $\lim _{x \rightarrow 2} \frac{x^{3}-8}{\sqrt{x}-\sqrt{2}}$.
View full question & answer→Question 43 Marks
Find the value of $\lim _{x \rightarrow 2} \frac{\sqrt{x+7}-3}{x-2}$.
View full question & answer→Question 53 Marks
Find the value of $\lim _{x \rightarrow 0} \frac{\sqrt{3+x}-\sqrt{3}}{x}$.
View full question & answer→Question 63 Marks
Find the value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\frac{2 x+3}{3 x-5}+\frac{3}{5}\right]$.
View full question & answer→Question 73 Marks
Find the value of $\lim _{x \rightarrow 1} \frac{2 x^{2}+x-3}{x^{2}-1}$.
View full question & answer→Question 83 Marks
Find the value of $\lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x^{2}-5 x+6}$.
Answer
$=\frac{3+1}{3-2}$
$=\frac{4}{1}$
$=4$
View full question & answer→$=\frac{3+1}{3-2}$
$=\frac{4}{1}$
$=4$
Question 93 Marks
Find the values of the following: $\lim\limits_{x \rightarrow 0} \frac{(1+x)^n-1}{x}$
Answer
View full question & answer→Put $1+x=t$.
So when $x \rightarrow 0$, then $t \rightarrow 1$ and $x=t-1$
$\therefore \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} =\lim _{t \rightarrow 1} \frac{t^n-1^n}{t-1}$
$ =n \cdot(1)^{n-1}$
$\left(\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n \cdot a^{n-1}\right)$
$ =n$
Hence, $\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}=n$
So when $x \rightarrow 0$, then $t \rightarrow 1$ and $x=t-1$
$\therefore \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} =\lim _{t \rightarrow 1} \frac{t^n-1^n}{t-1}$
$ =n \cdot(1)^{n-1}$
$\left(\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n \cdot a^{n-1}\right)$
$ =n$
Hence, $\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}=n$
Question 103 Marks
Find the values of the following: $\lim\limits_{x \rightarrow 0} \frac{\sqrt[10]{1+x-1}}{x}$
View full question & answer→Question 113 Marks
Find the values of the following: $\lim _{x \rightarrow 0} \frac{1}{x}\left[\frac{5 x+14}{3 x+7}-2\right]$
View full question & answer→Question 123 Marks
Find the values of the following: $\lim _{x \rightarrow-2} \frac{9 x^{2}+5 x-26}{5 x^{2}+17 x+14}$
View full question & answer→Question 133 Marks
Find the values of the following: $\lim _{x \rightarrow-\frac{1}{2}} \frac{2 x^{2}+3 x+1}{2 x^{2}-x-1}$
View full question & answer→Question 143 Marks
Find the values of the following: $\lim _{x \rightarrow-3} \frac{2 x^{2}+9 x+9}{2 x^{2}+7 x+3}$
View full question & answer→Question 153 Marks
Find the values of the following: $\lim _{x \rightarrow \frac{1}{2}} \frac{2 x^{2}+5 x-3}{4 x^{2}-1}$
View full question & answer→Question 163 Marks
Find the values of the following: $\lim _{x \rightarrow 1} \frac{x^{2}+2 x-3}{x^{2}-1}$
View full question & answer→Question 173 Marks
Find the values of the following: $\lim _{x \rightarrow-1} \frac{3 x^{2}-2 x-5}{x+1}$
View full question & answer→Question 183 Marks
Find the values of the following: $\lim _{x \rightarrow 3} \frac{x-3}{2 x^{2}-3 x-9}$
View full question & answer→Question 193 Marks
Find the values of the following: $\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1}$
View full question & answer→Question 203 Marks
Find the values of the following: $\lim _{x \rightarrow 1} \frac{x^{\frac{7}{2}}-1}{x^{\frac{3}{2}}-1}$
View full question & answer→Question 213 Marks
Find the values of the following :
$\lim _{x \rightarrow-1} \frac{x^{2017}+1}{x^{2018}-1}$
View full question & answer→$\lim _{x \rightarrow-1} \frac{x^{2017}+1}{x^{2018}-1}$
Question 223 Marks
Find the values of the following :
$\lim _{x \rightarrow-2} \frac{x^{10}-1024}{x^5+32}$
View full question & answer→$\lim _{x \rightarrow-2} \frac{x^{10}-1024}{x^5+32}$
Question 233 Marks
Find the values of the following: $\lim _{x \rightarrow 3} \frac{x^{6}-729}{x^{4}-81}$
View full question & answer→Question 243 Marks
Find the values of the following: $\lim _{x \rightarrow-P} \frac{x^{4}-p^{4}}{x^{3}+p^{3}}$
View full question & answer→Question 253 Marks
$\lim _{x \rightarrow 0} 1+\frac{2}{3+\frac{4}{x}}$
View full question & answer→Question 263 Marks
Find the values of the following: $\lim _{x \rightarrow 2}\left[\frac{2}{x-2}-\frac{4}{x^{2}-2 x}\right]$
View full question & answer→Question 273 Marks
Find the values of the following : $\lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+1}{x-1}$
View full question & answer→Question 283 Marks
If $y = 5 – 2x,$ show that as $x \rightarrow – 1$ then $y \rightarrow 7$ using tabular method.
Answer
View full question & answer→Here, $y = 5 – 2x.$ Taking the values of $x$ very close to $– 1,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $– 1$ by increasing or decreasing its value, the value of $y$ approaches to $7.$
That is, when $x \rightarrow – 1, y \rightarrow 7.$
Hence, it is proved that when $x \rightarrow – 1,$ then $y \rightarrow 7.$
It is clear from the table that when $x$ is brought nearer to $– 1$ by increasing or decreasing its value, the value of $y$ approaches to $7.$
That is, when $x \rightarrow – 1, y \rightarrow 7.$
Hence, it is proved that when $x \rightarrow – 1,$ then $y \rightarrow 7.$
Question 293 Marks
$\text { If } \mathrm{y}=\frac{x^2+x-6}{x-2} \text {, show that as } \mathrm{x} \rightarrow 2 \text {, then } \mathrm{y} \rightarrow 5 \text { using tabular method. }$
Answer
View full question & answer→$\text { Here, } y=\frac{x^2+x-6}{x-2}$
$ =\frac{x^2+3 x-2 x-6}{x-2}$
$ =\frac{(x-2)(x+3)}{x-2}$
$ =x+3$
$\text { We have to prove that, when } x \rightarrow 2 \text {, then } y \rightarrow 5 \text {. So we have to find } \lim _{x \rightarrow 2} x+3 \text {. }$
Here, $y = x + 3.$ Taking the values of $x$ very close to $2,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $2$ by increasing or decreasing its value, the value of $y$ approaches to $5.$ That is, when $x$ $\rightarrow 2, y \rightarrow 5.$
Hence, it is proved that when $x \rightarrow 2,$ then $y \rightarrow 5.$
$ =\frac{x^2+3 x-2 x-6}{x-2}$
$ =\frac{(x-2)(x+3)}{x-2}$
$ =x+3$
$\text { We have to prove that, when } x \rightarrow 2 \text {, then } y \rightarrow 5 \text {. So we have to find } \lim _{x \rightarrow 2} x+3 \text {. }$
Here, $y = x + 3.$ Taking the values of $x$ very close to $2,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $2$ by increasing or decreasing its value, the value of $y$ approaches to $5.$ That is, when $x$ $\rightarrow 2, y \rightarrow 5.$
Hence, it is proved that when $x \rightarrow 2,$ then $y \rightarrow 5.$
Question 303 Marks
$\text { Using tabular method, show that } \lim _{x \rightarrow 3} \frac{2}{x-3} \text { does not exist. }$
Answer
View full question & answer→Here, $f(x)=\frac{2}{x-3}$. Taking the values of $x$ very close to $3 ,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $3$ by increasing or decreasing its value, the value of $f(x)$ does not approaches to a definite value. That is, when $x \rightarrow 3, f(x)$ does not tend to a definite value. Hence, the limit of this function does not exist.
$\therefore \lim _{x \rightarrow 3} \frac{2}{x-3} \text { does not exist. }$
It is clear from the table that when $x$ is brought nearer to $3$ by increasing or decreasing its value, the value of $f(x)$ does not approaches to a definite value. That is, when $x \rightarrow 3, f(x)$ does not tend to a definite value. Hence, the limit of this function does not exist.
$\therefore \lim _{x \rightarrow 3} \frac{2}{x-3} \text { does not exist. }$
Question 313 Marks
Find the values of the following using tabular method: $\lim _{x \rightarrow 2} x$
Answer
View full question & answer→ Here, $f(x) = x.$ Taking the values of $x$ very close to $2,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $2$ by increasing or decreasing its value, the value of $f(x)$ approaches to $2.$
That is, when $x \rightarrow 2, f(x) \rightarrow 2.$
$\therefore \lim _{x \rightarrow 2} x=2$
It is clear from the table that when $x$ is brought nearer to $2$ by increasing or decreasing its value, the value of $f(x)$ approaches to $2.$
That is, when $x \rightarrow 2, f(x) \rightarrow 2.$
$\therefore \lim _{x \rightarrow 2} x=2$
Question 323 Marks
Find the values of the following using tabular method: $\lim _{x \rightarrow-3} \frac{2 x^{2}+9 x+9}{x+3}$
Answer
View full question & answer→$\text { Here, } f(x)=\frac{2 x^2+9 x+9}{x+3}$
$ =\frac{2 x^2+6 x+3 x+9}{x+3}$
$ =\frac{(2 x+3)(x+3)}{x+3}$
$ =2 x+3$
Taking the values of $x$ very close to $– 3,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $– 3$ by increasing or decreasing its value, the value of $f(x)$ approaches to $– 3.$
That is, when $x \rightarrow – 3, f(x) \rightarrow – 3$
$\therefore \lim _{x \rightarrow-3} \frac{2 x^2+9 x+9}{x+3}=-3$
$ =\frac{2 x^2+6 x+3 x+9}{x+3}$
$ =\frac{(2 x+3)(x+3)}{x+3}$
$ =2 x+3$
Taking the values of $x$ very close to $– 3,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $– 3$ by increasing or decreasing its value, the value of $f(x)$ approaches to $– 3.$
That is, when $x \rightarrow – 3, f(x) \rightarrow – 3$
$\therefore \lim _{x \rightarrow-3} \frac{2 x^2+9 x+9}{x+3}=-3$
Question 333 Marks
Find the values of the following using tabular method: $\lim _{x \rightarrow 2} \frac{2 x^{2}+3 x-14}{x-2}$
Answer
View full question & answer→$\text { Here, } f(x)=\frac{2 x^2+3 x-14}{x-2}$
$ =\frac{2 x^2+7 x-4 x-14}{x-2}$
$ =\frac{(x-2)(2 x+7)}{x-2}$
$ =2 x+7$
Taking the values of $x$ very close to $2, $ the following table is prepared :
It is clear from the table that when x is brought nearer to $2$ by increasing or decreasing its value, the value of $f(x)$ approaches to $11.$
That is, when $x \rightarrow 2, f(x) \rightarrow 11$
$\therefore \lim _{x \rightarrow 2} \frac{2 x^2+3 x-14}{x-2}=11$
$ =\frac{2 x^2+7 x-4 x-14}{x-2}$
$ =\frac{(x-2)(2 x+7)}{x-2}$
$ =2 x+7$
Taking the values of $x$ very close to $2, $ the following table is prepared :
It is clear from the table that when x is brought nearer to $2$ by increasing or decreasing its value, the value of $f(x)$ approaches to $11.$
That is, when $x \rightarrow 2, f(x) \rightarrow 11$
$\therefore \lim _{x \rightarrow 2} \frac{2 x^2+3 x-14}{x-2}=11$
Question 343 Marks
Find the values of the following using tabular method: $\lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3}$
Answer
View full question & answer→Here, $f(x)=\frac{x^2-2 x-3}{x-3}$
$=\frac{x^2-3 x+x-3}{x-3}$
$ =\frac{(x+1)(x-3)}{x-3}$
$ =x+1$
Taking the values of $x$ very close to $3,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $3$ by increasing or decreasing its value, the value of $f(x)$ approaches to $4.$
That is, when $x \rightarrow 3, f(x) \rightarrow 4.$
$\therefore \lim _{x \rightarrow 3} \frac{x^2-2 x-3}{x-3}=4$
$=\frac{x^2-3 x+x-3}{x-3}$
$ =\frac{(x+1)(x-3)}{x-3}$
$ =x+1$
Taking the values of $x$ very close to $3,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $3$ by increasing or decreasing its value, the value of $f(x)$ approaches to $4.$
That is, when $x \rightarrow 3, f(x) \rightarrow 4.$
$\therefore \lim _{x \rightarrow 3} \frac{x^2-2 x-3}{x-3}=4$
Question 353 Marks
Find the values of the following using tabular method: $\lim _{x \rightarrow 1} 2 x+1$
Answer
View full question & answer→Here, $f(x) = 2x + 1.$ Taking the values of $x$ very close to $1,$ the following table is prepared :
It is clear from the table that when $x$ is brought nearer to $1$ by increasing or decreasing its value, the value of $f(x)$ approaches to $3.$
That is, when $x \rightarrow 1, f(x) \rightarrow 3$
$\therefore \lim _{x \rightarrow 1} 2 x+1=3$
It is clear from the table that when $x$ is brought nearer to $1$ by increasing or decreasing its value, the value of $f(x)$ approaches to $3.$
That is, when $x \rightarrow 1, f(x) \rightarrow 3$
$\therefore \lim _{x \rightarrow 1} 2 x+1=3$
Question 363 Marks
Find the value of the following: $\lim _{x \rightarrow 3} \frac{x^{2}-3 x+5}{x+2}$
Answer
View full question & answer→$1$
Question 373 Marks
Find the value of the following: $\lim _{x \rightarrow a} \frac{x^{2}+a^{2}}{2 x a}$
Answer
View full question & answer→$1$
Question 383 Marks
Find the value of the following: $\lim _{x \rightarrow 5} \frac{x^{2}-3 x-10}{x+5}$
Answer
View full question & answer→$0$
Question 393 Marks
Find the value of the following : $\lim _{x \rightarrow 0} \frac{5 x^{2}+3 x-8}{x^{3}+8}$
Answer
View full question & answer→$-1$
Question 403 Marks
Find the value of the following: $\lim _{x \rightarrow k} \frac{x^{2}-\sqrt{x^{2}+3 k^{2}}}{x+k}$
Answer
View full question & answer→$\frac{k-2}{2}$
Question 413 Marks
Express the following interval forms In modulus form: $(1)\ (1,5)(2)(-1.5,1.5)(3)(-6.4,-3.6)$
Answer
View full question & answer→$(1)\ |x-3|<2(2)|x|<1.5(3)|x+5|<1.4$
Question 423 Marks
Express the following neighbourhood form in modulus form: $(1)\ N(-1,0.01)\ (2)\ N(-0.2,0.4)\ (3)\ N(5,2)$
Answer
View full question & answer→$(1)\ |x+1|<0.01 $
$(2)\ |x+0.2|<0.4$
$(3)\ |x-5|<2$
$(2)\ |x+0.2|<0.4$
$(3)\ |x-5|<2$
Question 433 Marks
Express the following modulus form in neighbourhood form: $(1)\ |x|<1\ (2)\ |2 x-1|<2\ (3)\ |x+2|<0.5$
Answer
View full question & answer→$(1)\ N(0,1)\ (2)\ N\left(\frac{1}{2}, 1\right)\ (3)\ N(-2,0.5)$