Qus : 2
1
lim , then find k
1
8/3
2
4/3
3
2/3
4
1
Go to Discussion
Solution
Quick DL Method Solution
Given:
\lim_{x \to 1} \frac{x^4 - 1}{x - 1}
= \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2}
LHS using derivative:
\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4
RHS using DL logic:
\lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2}
\approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2}
Equating both sides:
\frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3}
\boxed{k = \frac{8}{3}}
Qus : 3
3
Let f(x)=\frac{x^2-1}{|x|-1} . Then the value of lim_{x\to-1} f(x) is
1
-1
2
1
3
2
4
3
Go to Discussion
Solution
Qus : 5
4
The value of the limit \lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x} is
1
1
2
{3!}^{1/3!}
3
{3!}^{1/4}
4
{4!}^{1/4}
Go to Discussion
Solution
Qus : 6
4
Which of the following is NOT true?
1
\lim _{{x}\rightarrow\infty}\frac{x}{{e}^x}=0
2
\lim _{{x}\rightarrow0+}\frac{1}{{xe}^{1/x}}=0
3
\lim _{{x}\rightarrow0+}\frac{\sin x}{{1+2x}}=0
4
\lim _{{x}\rightarrow0+}\frac{\cos x}{{1+2x}}=0
Go to Discussion
Qus : 7
4
For a\in R (the set of al real numbers), a \ne 1 , \lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)(na+b)\ldots(na+n)\rbrack}=\frac{1}{60} . Then one of the value of a is
1
5
2
8
3
-15/2
4
-17/2
Go to Discussion
Qus : 8
3
The value of {{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x} is equal to
1
2
2
1
3
0
4
-1
Go to Discussion
Solution
Evaluate:
\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}
Step 1: Apply L'Hôpital's Rule (since it's 0/0):
First derivative:
\frac{e^x + e^{-x} - 2}{\sin x}
Still 0/0 → Apply L'Hôpital's Rule again:
\frac{e^x - e^{-x}}{\cos x}
Now,
\lim_{x \to 0} \frac{1 - 1}{1} = 0
Final Answer:
\boxed{0}
Qus : 9
1
\lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5} equal to
1
e^{5}
2
e^{-5}
3
e^{2}
4
e^{-2}
Go to Discussion
Solution
Qus : 10
1
Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If \lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3 , then f(2) is
1
0
2
4
3
- 8
4
- 4
Go to Discussion
Solution Given it has extremum values at x=1 and x=2
⇒f′(1)=0 and f′(2)=0
Given f(x) is a fourth degree polynomial
Let
f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e Given
\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3
\lim _{{x}\rightarrow0}\lbrack1+\frac{a{x}^4+b{x}^3+c{x}^2+\mathrm{d}x+e}{{x}^2}\rbrack=3
\lim _{{x}\rightarrow0}\lbrack1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{{x}^2}\rbrack=3
For limit to have finite value, value of 'd' and 'e' must be 0
⇒d=0 & e=0
Substituting x=0 in limit
⇒ c+1=3
⇒ c=2
f^{\prime}(x)=4a{x}^3+3b{x}^2+2cx+d
x=1 and x=2 are extreme values,
⇒f^{\prime}(1)=0 and $f^{\prime}(2)=0
⇒ 4a+3b+4=0 and 32a+12b+8=0
By solving these equations
we get, a=\frac{1}{2} and b=-2
So,
f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}
⇒f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)
⇒f(2)=0
Qus : 11
4
The value of \lim_{x\to a} \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}
1
\frac{2}{3}
2
\frac{2}{\sqrt{3}}
3
\frac{3\sqrt{3}}{\sqrt{2}}
4
\frac{2}{3\sqrt{3}}
Go to Discussion
Solution
Qus : 12
1
What is the value of \lim _{{x}\rightarrow\infty}-(x+1)\Bigg{(}{e}^{\frac{1}{x+1}}-1\Bigg{)} ?
1
-1
2
1
3
0
4
Does not exist
Go to Discussion
Solution
Qus : 13
4
1
1/2
2
-1/2
3
- 2
4
2
Go to Discussion
Solution Function is the form of
therefore using by L'Hospital rule
=
Again apply L'Hospital Rule,
=
Putting x = 0, we get
Qus : 14
2
Find
1
1
2
limit does not exist
3
infinity
4
None of these
Go to Discussion
Solution
Qus : 15
3
If f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)} is a real number then \lim _{{x}\rightarrow0}\, f(x)
1
2
2
3
3
\log _e2
4
\log _e3
Go to Discussion
Solution
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