Skip to content
EXERCISE 8.1 · Q24

Q.Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous: f(x)=e5sin⁡x−e2x5tan⁡x−3xf(x) = \dfrac{e^{5\sin x} - e^{2x}}{5\tan x - 3x}, for x≠0x \ne 0, =3/4= 3/4, for x=0x = 0, at x=0x = 0.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
33% · 24/73 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

f(x)=e5sin⁡x−e2x5tan⁡x−3xf(x)=\dfrac{e^{5\sin x}-e^{2x}}{5\tan x-3x} for x≠0x\ne0, f(0)=3/4f(0)=3/4.

Write the numerator as (e5sin⁡x−1)−(e2x−1)\big(e^{5\sin x}-1\big)-\big(e^{2x}-1\big) (subtracting and adding 11 does not change the difference). Dividing everything by xx:

e5sin⁡x−1x=e5sin⁡x−15sin⁡x⋅5sin⁡xx→1⋅5=5,e2x−1x→2,\frac{e^{5\sin x}-1}{x}=\frac{e^{5\sin x}-1}{5\sin x}\cdot\frac{5\sin x}{x}\to1\cdot5=5,\qquad \frac{e^{2x}-1}{x}\to2,

so the numerator divided by xx tends to 5−2=35-2=3. Similarly 5tan⁡x−3xx=5⋅tan⁡xx−3→5−3=2\dfrac{5\tan x-3x}{x}=5\cdot\dfrac{\tan x}{x}-3\to5-3=2.

lim⁡x→0f(x)=32.\lim_{x\to0} f(x)=\frac{3}{2}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.